Thermal Properties of Matter Class Notes PDF

# UMMEED NEET ## One Shot Physics ## Thermal Properties of Matter **By - Pawan Pandey Sir** **Physics Wallah** ### Topics to be covered: * **Thermal Expansion** * **Calorimetry** * **Temperature and Thermometrey** * **Heat transfer** * **Conduction** * **Convection** * **Radiation** #...

# UMMEED NEET ## One Shot Physics ## Thermal Properties of Matter **By - Pawan Pandey Sir** **Physics Wallah** ### Topics to be covered: * **Thermal Expansion** * **Calorimetry** * **Temperature and Thermometrey** * **Heat transfer** * **Conduction** * **Convection** * **Radiation** ### Thermal Expansion #### Linear Expansion The image shows a diagram of a rod with an initial length *l*. When the temperature increases by *ΔT*, the length of the rod increases by *Δl*. The coefficient of linear expansion is represented by *∝* and is defined as the change in length per unit length per degree Celsius. The following formulas summarize the concepts of linear expansion: * *Δl= l∝ΔT* * *l’ = l + ΔT* * *l’ = l(1 + ∝ΔT)* #### Areal Expansion The image shows a diagram of a rectangular plate with an initial area *A*. When the temperature increases by *ΔT*, the area of the plate increases by ΔA*. The coefficient of areal expansion is represented by *β* and is defined as the change in area per unit area per degree Celsius. The following formulas summarize the concepts of areal expansion: * *ΔA = AβΔT* * *β =2∝* (Isotropic Body) * *β =∝₁ + ∝₂* (Anisotropic Body) * *Α’=Α(1+BAT)* #### Volumetric Expansion The image shows a diagram of a cube with an initial volume *V*. When the temperature increases by *ΔT*, the volume of the cube increases by *ΔV*. The coefficient of volumetric expansion is represented by *γ* and is defined as the change in volume per unit volume per degree Celsius. The following formulas summarize the concepts of volumetric expansion: * *ΔV = VYΔT* * *γ =3∝* (Isotropic Body) * *γ =∝₁ + ∝₂ +∝₃* (Anisotropic Body) * *V’= V(1 + YΔT)* ### Density The image shows a diagram of a cube with an initial volume *V*. When the temperature increases by *ΔT*, the volume of the cube increases by *ΔV*. This results in a change in density. The following formulas summarize the concepts of density. * *ρ = m/V'* * *ρ’= m/V’ = m/(V(1 + YΔT)) = ρ/(1 + YΔT)* * *ρ’ = ρ(1 - YΔT)* ### Bi-metallic Strip The image shows a diagram of a bimetallic strip made of two different metals with different coefficients of linear expansion. When the temperature increases, the metal with the higher coefficient of linear expansion expands more than the metal with the lower coefficient. This difference in expansion causes the bimetallic strip to bend. ### Photographic Expansion The image shows a diagram of a yellow rectangle with a circular hole in the center. When the temperature increases, the rectangle expands, but the hole gets smaller because the material surrounding the hole expands outward. ### Thermal Stress The image shows a diagram of a rod that is fixed at one end and is subjected to an increase in temperature. The rod tries to expand, but the fixed end prevents it from doing so. This creates a stress in the rod. The following formulas summarize the concept of thermal stress. * *Δl = l∝ΔT* * *F = YΔl/A = YA∝ΔT/A* * *Thermal stress = Y X strain = Y X (Δl/l) = Y∝ΔT* ### Pendulum The image shows a diagram of a simple pendulum. When the temperature increases, the length of the pendulum increases, which affects its period. The hotter the temperature, the slower the pendulum swings. The following are formulas that incorporate this concept. * *T = 2π√(L/g)* * *ΔT = (∝ΔT)/2* ### Series Combination of Rods The image shows a diagram of two rods connected in series. The rods have different lengths (*l₁* and *l₂*) and different coefficients of linear expansion (*∝₁* and *∝₂*). The rods are heated by a temperature *ΔT*, and the image shows the formulas for the equivalent coefficient of linear expansion (*∝eq*) and the change in the equivalent length (*Δleq*). ### Question #1 #### *Prompt:* A steel rod with y = 2.0 x 10¹¹ Nm^-2 and ∝ = 10^-5 °C^-1 of length 4 m and area of cross-section 10 cm² is heated from 0°C to 400°C without being allowed to expand. The tension produced in the rod is x × 10⁵ N where the value of x is #### *Solution:* * *F = YA∝ΔT = 2x10¹¹ x (10x10^-4) x 10^-5 x (400 - 0) = 8x10⁵* Therefore, the value of *x* is **8**. ### Question #2 #### *Prompt:* A non-isotropic solid metal cube has coefficients of linear expansion as : 5 x 10^-5/°C along the x-axis and 5 × 10^-6/°C along the y and the z -axis. If the coefficient of volume expansion of the solid is C 10^-6 /°C then the value of C is #### *Solution:* * *γ = ∝ₓ + ∝ᵧ + ∝_z = 5x10^-5 + 2(5x10^-6) = 5x10^-5 + 10^-5 = 6x10^-5 = 60x10^-6* Therefore, the value of *C* is **60**. ### Question #3 #### *Prompt:* Two different wires having lengths L₁ and L₂, and respective temperature coefficient of linear expansion ∝₁, and ∝₂, are joined end-to-end. Then the effective temperature coefficient of linear expansion is: #### *Solution:* * *∝_eq = (∝₁L₁ + ∝₂L₂)/(L₁ + L₂)* The answer is **4.** ### Question #4 #### *Prompt:* If two rods of lengths L and 2 L having coefficient of linear expansion ∝ and 2 ∝ respectively are connected end-on-end, the average coefficient of linear expansion of the composite rod equals #### *Solution:* * The lengths of the rods are *L₁ = L* and *L₂ = 2L*. * The coefficients of linear expansion of the rods are *∝₁ = ∝* and *∝₂ = 2∝*. * *∝_eq = (∝₁L1 + ∝₂L₂)/(L₁ + L₂) = ∝(L) + 2∝ (2L)/(L + 2L) = 5∝/3* The answer is **3.** ### Question #5 #### *Prompt:* A copper rod of 88 cm and an aluminium rod of unknown length *L₂* has difference in length independent of increase in temperature. The length of aluminium rod is: (∝_cu = 1.7 × 10^-5K^-1 and ∝_ai = 2.2 × 10^-5K^-1) #### *Solution:* * *L₂ - L₁ = constant*  * *L₂∝_aiΔT = L₁∝_cuΔT* * *L₂ = (L₁∝_cu)/∝_ai = (88 x 1.7 x 10^-5)/(2.2 x 10^-5) = 68cm* The answer is **2.** ### Question #6 #### *Prompt:* A metallic bar is heated from 0°C to 100°C. The coefficient of linear expansion is 10^-5 K^-1. What will be the percentage increase in length: #### *Solution:* * *Δl = l∝ΔT* * *% increase = (Δl/l) x 100 = ∝ΔT x 100* * *% increase = 10^-5 x (100-0) x 100 = 0.1%* The answer is **2**. ### Latent Heat The image shows that the phase change from solid to liquid requires a certain amount of heat. This heat is called the *latent heat of fusion*. The phase change from liquid to gas requires a certain amount of heat. This heat is called the *latent heat of vaporization*.  * *Q = mLf* for the phase change from solid to liquid. * *Q = mLv* for the phase change from liquid to gas. ### Heat Capacity The image shows that the amount of heat required to raise the temperature of an object by a certain amount is called the *heat capacity* of the object. The heat capacity is proportional to the mass of the object. * *C = mS* * *Q = CΔT = mSΔT* ### Principle of Calorimetry The heat loss of a system in equilibrium equals the heat gain of the system.  * *Heat Loss = Heat Gain* ### Question #7 #### *Prompt:* A heater gives heat at rate of 200 W. It is used to heat 50 g water from 20°C to 40°C. Assuming no heat loss, the time taken is: #### *Solution:* * *Q = mSΔT = 50g x 1cal/g°C x (40°C - 20°C) = 1000cal = 4200J*  * *P = Q/t = 200J/s* * *t = Q/P = 4200J/(200J/s) = 21s* The answer is **3.** ### Question #8 #### *Prompt:* The quantities of heat required to raise the temperature of two solid iron spheres of radii r₁ and r₂(r₁ = 3r₂) by 100K are in the ratio #### *Solution:* * *Q = mSΔT* and *m ∝ volume* * *Q₁ = (4/3)πr₁³ρSΔT* * *Q₂ = (4/3)πr₂³ρSΔT* * *Q₁/Q₂ = (r₁/r₂)³= (3r₂/r₂)³= 27* The answer is **1.** ### Question #9 #### *Prompt:* A vessel of copper of mass 0.5 kg contains 30 g of water at 80°C. If 100 g water at 10°C is poured into it, find the final temperature of mixture. (Specific heat capacity of water = 4.2 J/g°C; Specific heat capacity of of Copper = 0.4 J/g°C) #### *Solution:* * *Heat Loss = Heat Gain* * *m_cuS_cu(80 - T) + m₁S₁(80 - T) = m₂S₂(T - 10)*  * *500 x 0.4 x (80 - T) + 30 x 4.2(80 - T) = 100 x 4.2(T - 10)* * *T = 28.5°C* ### Question #10 #### *Prompt:* A piece of ice falls from a height *h* so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of *h* is (Latent heat of ice is 3.4 × 10⁵ J/kg and g = 10 N/kg) #### *Solution:* * *U = mgh = (1/4)mLf* * *h = (m/4)Lf/mg = (1/4)(3.4 x 10⁵ J/kg)/ 10N/kg = 8500m = 8.5km* The answer is **3.** ### Question #11 #### *Prompt:* 10 g of ice cubes at 0°C are released into a beaker (water equivalent 55 g) at 40°C. Assuming that negligible heat is taken from the surrounding, the temperature of water in the beaker becomes nearly #### *Solution:* * *Heat Gain = Heat Loss* * *mLf + mSΔT = M_bS_bΔT* * *10g x 80cal/g + 10g x 1cal/g°C x (T - 0°C) = 55g x 1cal/g°C x (40°C - T)* * *T = 22°C* The answer is **2.** ### Question #12 #### *Prompt:* If 1 g of steam is mixed with 1 g of ice, then resultant temperature of the mixture is #### *Solution:* * The steam will condense to water at 100°C, and then the water will cool down to 0°C. The ice will melt to water at 0°C. The final temperature of the mixture will be 0°C because the heat released by the steam condensing and cooling is less than the heat required to melt the ice. The answer is **2**. ### Question #13 #### *Prompt:* A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at #### *Solution:* * *T₁λ₁ = T₂λ₂* * *λ₂ = (T₁λ₁)/T₂ = (1500K x 5000Å)/(2500K) = 3000Å* The answer is **1**. ### Question #14 #### *Prompt:* On observing light from three different stars P, Q and R, it was found that intensity of violet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the spectrum of R, and the intensity of red colour is maximum in the spectrum of Q. If T_P, T_Q, and T_R are the respective absolute temperatures of P, Q and R then it can be concluded from the above observations that #### *Solution:* * The wavelength of violet light is shorter than the wavelength of green light, which is shorter than the wavelength of red light. * According to Wien’s displacement law, *λ_mT = constant* * Therefore, *T_P > T_R > T_Q.* The answer is **2**. ### Question #15 #### *Prompt:* Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is #### *Solution:* * *P = σeAT⁴* * *P₁ / P₂ = (σeA₁T₁⁴) / (σeA₂T₂⁴) = [(4πr₁² T₁⁴)] / [(4πr₂²T₂⁴)] = (r₁²T₁⁴)/(r₂²T₂⁴) = (1² x 4000⁴)/(4² x 2000⁴) = 1* The answer is **1**. ### Question #16 #### *Prompt:* A spherical black body with a radius of 12 cm radiates 450 watt power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be: #### *Solution:* * *P = σeAT⁴ = σe4πr²T⁴* * *Let P₁ = 450 Watt* * *If r₂= (1/2)r₁ and T₂ = 2T₁, then P₂ = σ, *e*4π(r₂)² *(T₂)⁴ = σ, *e*4π( (1/2)r₁)²*(2T₁)⁴ = (1/4) σ, *e*4π(r₁)²*(16T₁⁴) = 4σ, *e*4π(r₁)²*(T₁)⁴ = 4P₁ = 4 x 450 Watt = 1800 Watt* The answer is **4**. ### Question #17 #### *Prompt:* The power radiated by a black body is P and it radiates maximum energy at wavelength *λ_o*. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength (3/4) *λ_o*, the power radiated by it becomes nP. The value of n is #### *Solution:* * *P ∝ T⁴ ∝ 1/λ_m* * *T₁λ_m1= T₂λ_m2* * *(3/4)λ_o x T₂ = λ_o x T₁* * *T₂ = (4/3)T₁* * *P₂/P₁ = (T₂/T₁)⁴ = [(4/3)T₁/T₁]⁴ = (4/3)⁴ = 256/81* * *P₂ = (256/81)P₁ = nP₁* The answer is **3**. ### Question #18 #### *Prompt:* Assuming the sun to have a spherical outer surface of radius *r*, radiating like a black body at temperature *t* , the power received by a unit surface,(normal to the incident rays)at a distance R from the centre of the sun is #### *Solution:* * *I = (P/A) = (σeAT⁴) / (4πR²) = σr²(t+273)⁴ / R²* The answer is **3**. ### Question #19 #### *Prompt:* An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80°C to 70°C. The time taken to cook for the same object from 70°C to 60°C would be nearly #### *Solution:* * According to Newton's Law of Cooling, *—(ΔT/Δt) ∝ (T - T_s)*. * *We can say that (T₁ - T_s) / t₁ = (T₂ - T_s) / t₂*  * *If T₁ = 80°C, T₂ = 70°C, T_s = 25°C, then (80-25)/12 = (70-25)/t₂*. * *t₂ = 7.5 min*. * *Repeating the process one more time with T₁ = 70°C, and t₁ = 7.5min, we can find the time it takes to cool to 60°C*. * *For this situation: (70-25)/7.5 = (60-25)/t₂*. * *t₂ = 5 min.* The answer is **1**. ### Question #20 #### *Prompt:* Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is #### *Solution:* * According to Newton's Law of Cooling, *—(ΔT/Δt) ∝ (T - T_s)*. * *This implies that the rate of cooling is proportional to the **temperature difference** between the object and the surroundings*. * *In the first 5 minutes, the water cools by 10°C*. * *In the next 5 minutes, the water cools by 6°C*. * *Thus, the rate of cooling is **decreasing**, indicating that the temperature difference between the water and the surroundings is decreasing*. * *Therefore, the temperature of the surroundings is **lower than the temperature of the water at the beginning of each 5 minute interval***. * *More specifically, it must be lower than the average temperature of the water in that interval*. * *The average temperature of the water during the first interval is (70+60)/2 = 65°C*. * *The average temperature of the water during the second interval is (60+54)/2 = 57°C*. * *This tells us that the temperature of the surroundings must be closer to 57°C than it is to 65°C*. * *Of the choices provided, only 42°C is the temperature of the surroundings*. The answer is **3**. ### Question #21 #### *Prompt:* A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be #### *Solution:* * *(T₁-T_s)/t₁ = (T₂-T_s)/t₂* * *(3T-T)/10 = (2T-T)/10* This implies that the temperature of the body will cool by *T* in the next 10 minutes. So, the body's temperature will be 2T - T = **T** at the end of the next 10 minutes. The answer is **1**. ### Question #22 #### *Prompt:* Two rods with the same dimensions have thermal conductivities in the ratio 1: 2. They are arranged between heat reservoirs with the same temperature difference, in two different configurations, A and B. The rates of heat flow in A and B are I_A and I_B respectively. The ratio I_A/I_B is equal to #### *Solution:* * *For configuration A, the rods are in parallel*. * *For configuration B, the rods are in series*. * *The equivalent thermal conductivity of the system in parallel is K_eq =(K₁A₁ + K₂A₂) / (A₁ + A₂) = (K+2K) / (1+1) = 3/2K*. * *The equivalent thermal conductivity of the system in series is K_eq = 2L/[(L/K) +(L/2K)] = 4/3K* * *Therefore, for the system in parallel, I_A=(ΔT/R) = ΔT(K_eqA/L) = ΔT(3/2K)(A/L) = 3ΔTA/2KL*. * *For the system in series, I_B=(ΔT/R) = ΔT(K_eqA/L) = ΔT(4/3K)(A/L) = 4ΔTA/3KL*. * *Therefore, I_A/I_B = [3ΔTA/2KL] / [4ΔTA/3KL] = 9/8* The answer is **4**. ### Question #23 #### *Prompt:* The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x respectively are T₂ and T₁(T₂> T₁). The rate of heat transfer through the slab, in a steady state is [(A(T₂-T₁)K)/x] with *f* equal to: #### *Solution:* * *L₁ =x and L₂ = 4x* * *K₁ = K and K₂ = 2K* * *R₁ = L₁/(K₁A) = x/KA and R₂ = L₂/(K₂A) = 4x/2KA = 2x/KA* * *R_eq = R₁ + R₂ = x/KA + 2x/KA = 3x/KA*. * *i = (T₂ -T₁)/R_eq = (A(T₂ -T₁)K)/3x* The answer is **4**. ### Question #24 #### *Prompt:* For a black body at temperature 727°C, its radiating power is 60 watt and temperature of surrounding is 227°C. If temperature of black body is changed to 1227°C then its radiating power will be: #### *Solution:* * *P = σeAT⁴* * *As the area and emissivity remain constant*: * *P₁/P₂ = (T₁)⁴/(T₂)⁴ = (1000)⁴/(1500)⁴ =(2/3)⁴ = 16/81* * *Since P₁ = 60 Watts, then P₂ = (P₁ x 81)/16 = (60 Watts x 81)/16 = 303.75 Watts* The answer is **1**. ### Question #25 #### *Prompt:* The coefficient of thermal conductivity of copper is nine times that of steel in the composite cylindrical bar shown in the figure. The temperature at the junctions of copper & steel is approximately equal to? #### *Solution:* * *The rate of heat flow through the copper rod is equal to the rate of heat flow through the steel rod*, since they are in series. * *i_Cu = i_steel* * *The thermal conductivity of copper is 9 times that of steel (K_Cu = 9K_steel)*. * *The lengths of the rods are L_Cu = 20 cm* and *L_steel = 10 cm*. * *Therefore, (K_CuA(80-T)/L_Cu) = (K_steelA(T-0)/L_steel)*. * *Simplifying, (9K_steelA(80-T)/20) = (K_steelA(T)/10) => 9(80-T) = 2T => 720 - 9T = 2T => 11T = 720* * *Therefore, T = 720/11 = 65.45°C* The answer is **3**.