Physical Therapy Lecture 4 2024-2025 PDF

Lecture no. 4 Biophysics MED 106 1st semester 2024-2025 Physical Therapy Students Prof. El-shazly M. Duraia [email protected] https://www.researchgate.net/profile/El-Shazly_Duraia Measurements of the specific heat of water, The...

Lecture no. 4 Biophysics MED 106 1st semester 2024-2025 Physical Therapy Students Prof. El-shazly M. Duraia [email protected] https://www.researchgate.net/profile/El-Shazly_Duraia Measurements of the specific heat of water, Therapeutic heat and cold in medicine, Application of heat and cold. Dr. El-shazly M. Duraia [email protected] https://www.researchgate.net/profile/El-Shazly_Duraia Determination of the Specific Heats A calorimeter is a lightweight, insulated flask containing water. When an object is put in, it and the water come to thermal equilibrium. If the mass of the flask can be ignored, and the insulation keeps any heat from escaping: 1. The final temperatures of the object and the water will be equal. 2. The total energy of the system is conserved. This allows us to calculate the specific heat of the object. What will happen over time? Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291  Let’s take a closer look… Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291 Eventually, the temperatures will equalize  Thermometer Much calorimetry is carried out using a coffee-cup calorimeter, under constant pressure Styrofoam (i.e. atmospheric pressure) cover Styrofoam If we assume that no heat is cups lost to the surroundings, then the energy absorbed inside the calorimeter must be equal to the Stirrer energy released inside the calorimeter. i.e., q absorbed = – q released qx = – qy  1. A 75.0 g piece of lead (specific heat = 0.130 J/goC), initially at 435oC, is set into 125.0 g of water, initially at 23.0oC. What is the final temperature of the mixture? Pb What is the final 75.0 g 125 g temperature, Tf, of 435.0 °C 23.0 °C the mixture? C = 0.130 J/°C g qwater = –qPb q = m x C x ΔT for both cases, although specific values differ Plug in known information for each side Solve for Tf... A 75.0 g piece of lead (specific heat = 0.130 J/goC), initially at 435oC, is set into 125.0 g of water, initially at 23.0oC. What is the final temperature of the mixture? q = m x C x ΔT for both cases, although specific values differ Plug in known information for each side qwater = –qPb mwater Cwater DTwater = –mPb CPb DTPb 125 (4.18) (Tf – 23) = –75 (0.13) (Tf – 435) = Tf = 30.5oC 2. A 97.0 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15.0oC. If gold has a specific heat of 0.129 J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. T = 785oC Au mass = 97.0 g T = 15.0 oC mass = 323 g - LOSE heat = GAIN heat - [(C Au) (mass) (DT)] = (C H2O) (mass) (DT) - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] - [(12.5) (Tf - 785oC)] = (1.35 x 103) (Tf - 15oC)] -12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104 3 x 104 = 1.36 x 103 Tf Tf = 22.1oC HW #2. If 59.0 g of water at 13.0 oC are mixed with 87.0 g of water at 72.0 oC, find the final temperature of the system. T = 13.0 oC T = 72.0 oC mass = 59.0 g mass = 87.0 g - LOSE heat = GAIN heat - [ (mass) (C H2O) (DT)] = (mass) (C H2O) (DT) - [ (59 g) (4.184 J/goC) (Tf - 13oC)] = (87 g) (4.184 J/goC) (Tf - 72oC)] - [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)] -246.8 Tf + 3208 = 364 Tf - 26208 29416 = 610.8 Tf Tf = 48.2oC  HW #4. 240. g of water (initially at 20.0oC) are mixed with an unknown mass of iron initially at 500.0oC (CFe = 0.4495 J/goC). When thermal equilibrium is reached, the mixture has a temperature of 42.0oC. Find the mass of the iron. T = 500oC Fe mass = ? grams T = 20oC mass = 240 g - LOSE heat = GAIN heat -q1 = q2 - [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT) - [ (X g) (0.4495 J/goC) (42oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20oC)] - [ (X) (0.4495) (-458)] = (240 g) (4.184) (22) 205.9 X = 22091 X = 107 g Fe  Cold therapy Painful arthritis inflammation can be treated with a cold compress. This is an inexpensive, effective treatment that can be used many times a day or occasionally, as needed Cold therapy can: Decrease inflammation. Cold causes the blood vessels of the muscles to constrict, and can therefore decrease the flow of blood and help reduce inflammation. Slow the production of joint fluid. Synovial joint fluid is essential to a healthy joint, but too much can contribute to swelling and discomfort. Applying a cold compress to a resting joint can slow the production of joint fluid. Distract the brain from the inflammation. Cooling an inflamed joint can stimulate sensory receptors in the skin and decrease the transmissions of pain signals to the brain.  Cold therapy Cold therapy constricts the blood vessels in the muscles. This constriction decreases blood flow to the affected area and helps to reduce inflammation. How long should cold be applied? How long should cold be applied? An ice or cold pack may be applied for no more than 20 minutes at a time, but this process can be repeated throughout the day—up to 8 or 10 times in a 24-hour period.  Heat Therapy: Heat has two primary therapeutic effects: 1. Increase in metabolism resulting in relaxation of the blood capillaries (vasodilation). 2. Increase in blood supply to cool down the heated area. Methods of producing heat in the body: 1. Conductive method 2. Radiation IR 3. Radio wave heating (diathermy) 4. Ultrasonic heating  Heat Therapy: ‘Diathermy' means 'through heating' or producing deep heating directly in the tissues of the body. Externally applied sources of heat like hot towels, infrared lamps and electric heating pads often produce discomfort and skin burns long before adequate heat has penetrated to the deeper tissues. In diathermy technique, the subject's body becomes a part of the electrical circuit and the heat is produced within the body and not transferred through the skin  Ultrasonic heating These waves are different from electromagnetic waves. It produces : mechanical vibration inside tissue. It is the same as the sound waves but it has much higher frequencies about 1MHz with power of several watts per centimeter. It can move the tissue particles backward and forward with high frequency, in doing so it can increase the kinetic energy consequently it heats the tissue.  Example: Suppose a person wish to lose 4.54 kg through a physical activity, how long would the person have to work at an activity of 15 kcal/min to lose 4.54kg of fat, the person can expect a maximum of 9.3kal/g of fat. Solution: the time required = (4.54x10^3g)(9.3kcal/g)/ 15 kcal/min = 2810 min = 47 h Please note that a great deal of exercise is required to lose few kilograms of fat  Example: Suppose a person wish to lose 4.54 kg through dieting, it is usually much easier to lose weight by reducing the food intake a person normally use 2500kcal/day how long must he diet to lose 4.54kg?. Solution: 4540 x 9.3 = 42222 kcal the time required = 42222/500 = 84.4 day  Determination of BMR BMR is determined from the oxygen consumption at rest The amount of energy used by various organs can be determined form its oxygen consumptions They can also estimate the food energy used in various physical activities by measuring the oxygen consumption Carbon dioxide production and Heat production Determination of BMR Importance of BMR Body Mass Index (BMI) is an indicator of body fatness relative to a person’s height and weight. This measurement itself is not a diagnostic of the health of an individual but can be a pointer that tells the risk level of an individual to some health problems. BMI ranges from underweight to obese. Body Mass Index (BMI)