Heat gained by soda = Heat lost by watermelon Qsoda = Qwatermelon

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1 PHYS1114 SAMPLE EXAM 5 SOLUTIONS Spring 2013 Professor Kenny L. Tapp 1. Dermatologists often remove small precancerous skin lesions by freezing them quickly with liquid nitrogen, which has a temperature of 77 K. What is this temperature on the (a) Celsius and (b) Fahrenheit scales? a. The Kelvin temperature and the temperature on the Celsius scale are related by Equation 12.1: T = T c , where T is the Kelvin temperature and is the Celsius temperature. Therefore, a temperature of 77 K on the Celsius scale is b. The temperature of 196 C is 196 Celsius degrees below the ice point of 0 C. Since, this number of Celsius degrees corresponds to Subtracting 353 Fahrenheit degrees from the ice point of 32.0 F on the Fahrenheit scale gives a Fahrenheit temperature of. 2. An ice chest at a beach party contains 12 cans of soda at 5.0 C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg?C ). Someone adds a 6.5-kg watermelon at 27 C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon. We assume that no heat is lost through the chest to the outside. Then, energy conservation dictates that the heat gained by the soda is equal to the heat lost by the watermelon in reaching the final temperature. Each quantity of heat is given by the equation,, where we write the change in temperature as the higher temperature minus the lower temperature. Starting with the statement of energy conservation, we have Heat gained by soda = Heat lost by watermelon Qsoda = Qwatermelon (cmδt) soda = (cmδt) watermelon Since the watermelon is being treated like water, we take the specific heat capacity of water. Thus, the above equation becomes Solving for, we obtain 1

2 3. b) How much heat in kilocalories is required to change 2.3 kg of water from 38 C to 45 C? 4. When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37 C) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to burn calories? Suppose you expend 430 kilocalories during a brisk hour-long walk. How many liters of ice water (0 C) would you have to drink in order to use up 430 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter. The volume V of a mass m of water is given by, where r is the mass density of water. In order to warm a mass m of ice water to body temperature, the body must provide an amount Q of heat given by, where c is the specific heat of water, and ΔT is 37 C, the difference between body temperature (37 C) and the temperature of ice water (0 C). We will calculate the required mass m of ice water, and then find the corresponding volume V. The volume V of the water is & solving for the mass m, we obtain The amount Q of heat is given as 430 kcal, which must be converted to joules via 1 kcal = 4186 J: Solve for the volume of ice water in m 3 : To convert this result to liters, we use the equivalence 1 liter = : 2

3 5. A 20 g iron nail is being pounded by a 0.44 kg hammer. The speed of the hammer as it strikes the nail is 9.0 m/s. Assuming half the kinetic energy of the hammer is converted into thermal energy of the nail, how many times should the nail be stuck to raise its temperature by 17 C? (specific heat of iron = 448 J/kg C). 3

4 6. Find the total heat required to change 2.0 kg of ice at -10 C into water at 20 C. (Specific heat for ice = 2090 J/kg C, Latent Heat of Fusion = 3.34 x 105 J/kg, Latent Heat of Vaporization = 22.6 x 105 J/kg). 7. Define and provide an example for each of the three methods of heat transfer. CONDUCTION CONVECTION - RADIATION 4

5 8. Suppose that the steel gas tank in your car is completely filled when the temperature is 17 C. How many gallons will spill out of the twenty-gallon tank when the temperature rises to 35 C? Both the gasoline and the tank expand as the temperature increases. The coefficients of volumetric expansion β g and β s for gasoline and steel can be found in any standard table of data. According to the equation, the volume expansion of the gasoline is while the volume of the steel tank expands by an amount The amount of gasoline which spills out is 9. One early May morning during finals week at the Oklahoma City Community College, the temperature fell from 12.0 C to C in 27.0 minutes. Suppose that a 19-m aluminum flagpole were subjected to this temperature change. Find the average speed at which its height would decrease, assuming the flagpole responded instantaneously to the changing temperature. The average speed of the flagpole s contraction is given by, where ΔL is the amount by which it contracts, and Δt is the elapsed time. The amount of contraction the pole undergoes is found from, where a is the coefficient of thermal expansion for aluminum, L 0 is the length of the pole before it begins contracting, and ΔT is the difference between the higher and lower temperatures of the pole. Substituting into, we get The elapsed time Δt is given in minutes, which must be converted to seconds: Substituting this result and the given values into our master equation, we obtain END OF SAMPLE EXAM 5 5