A solid of mass 100 g at a temperature of 90°C is placed in 100g of water at 20°C in a container of negligible heat capacity. If the final steady temperature is 60°C, calculate the... A solid of mass 100 g at a temperature of 90°C is placed in 100g of water at 20°C in a container of negligible heat capacity. If the final steady temperature is 60°C, calculate the specific heat capacity of the solid. The specific heat capacity of water equals 4.2x10³ jkg–¹K–¹.
Understand the Problem
The question is a calorimetry problem. It asks to calculate the specific heat capacity of a solid, given the masses and initial temperatures of the solid and water, and the final equilibrium temperature. We will use the principle of conservation of energy, where the heat lost by the solid is equal to the heat gained by the water. Knowing the specific heat capacity of water, we can solve for the specific heat capacity of the solid.
Answer
$c_s = 0.10465 \, \text{J/g}^\circ\text{C}$
Answer for screen readers
$c_s = 0.10465 , \text{J/g}^\circ\text{C}$
Steps to Solve
- Define variables and write down known values
$m_s$ = mass of solid = 200 g $T_{s,i}$ = initial temperature of solid = 85 °C $m_w$ = mass of water = 100 g $T_{w,i}$ = initial temperature of water = 22 °C $T_f$ = final temperature of mixture = 25 °C $c_w$ = specific heat capacity of water = 4.186 J/g°C $c_s$ = specific heat capacity of solid = ? J/g°C
- Write the heat exchange equation
Heat lost by solid = Heat gained by water $Q_s = -Q_w$ $m_s c_s (T_f - T_{s,i}) = -m_w c_w (T_f - T_{w,i})$
- Plug in the values
$200 \cdot c_s \cdot (25 - 85) = -100 \cdot 4.186 \cdot (25 - 22)$
- Simplify the equation
$200 \cdot c_s \cdot (-60) = -100 \cdot 4.186 \cdot (3)$ $-12000 \cdot c_s = -1255.8$
- Solve for $c_s$
$c_s = \frac{-1255.8}{-12000}$ $c_s = 0.10465$ J/g°C
$c_s = 0.10465 , \text{J/g}^\circ\text{C}$
More Information
The specific heat capacity of the solid is 0.10465 J/g°C. This value indicates how much energy is required to raise the temperature of 1 gram of the solid by 1 degree Celsius.
Tips
A common mistake is forgetting the negative sign in the heat exchange equation. Remembering that heat lost is negative and heat gained is positive is crucial to avoid errors.
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