Problem

. A thermos for liquid helium. A physicist uses a…

05:23

Problem 86 Hard Difficulty

. Basal metabolic rate. The energy output of an animal engaged in an activity is called the basal metabolic rate (BMR) and is a measure of the conversion of food energy into other forms of energy. A simple calorimeter to measure the BMR consists of an insulated box with a thermometer to measure the temperature of the air. The air has a density of 1.29 $\mathrm{kg} / \mathrm{m}^{3}$ and a specific heat capacity of 1020 $\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .$ A 50.0 $\mathrm{g}$ hamster is placed in a calorimeter that contains 0.0500 $\mathrm{m}^{3}$ of air at room temperature. (a) When the hamster is running in a wheel, the temperature of the air in the calorimeter rises 1.8 $\mathrm{C}^{\circ}$ per hour. How much heat does the running hamster generate in an
hour? (Assume that all this heat goes into the air in the calorimeter. Neglect the heat that goes into the walls of the box and into the thermometer, and assume that no heat is lost to the surroundings.) (b) Assuming that the hamster converts seed into heat with an efficiency of 10$\%$ and that hamster seed has a food energy value of 24 $\mathrm{J} / \mathrm{g}$ , how many grams of seed must the hamster eat per hour to supply the energy found in part (a)?

Answer

a) 118 J/h
b) 49 $\mathrm{g} / \mathrm{h}$

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Watch More Solved Questions in Chapter 14

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Problem 14

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Video Transcript

everyone. This is question number 86 from Chapter 14. In this problem, we're talking about a hamster in a kalorama ter basil mint of metabolic rate and stuff. But so we have this hamster in Kalorama ter, and he's running on a wheel were told, were given the air density, the heat capacity of the air. The were given the mass of the Hampshire 50 grams, and we're told that the crown barometer contains your 500.5 Weiner's cubed ofher at room temperature, or then given that while the hamsters running on a wheel, the air temperature increases 1.8 degrees Celsius per hour. The Hampshire converts seed to heat a 10 degrees, 10% efficiency, and the seed has an energy value of 24 Jules program. So part a asked, asked us to find the heat generated in in an hour, so he generated in an hour. We don't have a face change or anything. We're just dealing with a change in temperature so we can use our basic equation. Q E equals M C. Delta team Kalorama Tree specifically should also remind you of this so we don't have the mass of the air but we have. We have density and volume and density equals mass over volume. So then mass equals Roe V, so then we can replace them with Robie. So then we have Q equals ro V C, which we have Delta T, which we have, so we can just plug these numbers in. So this gives us 1.29 kilograms per meter. Cute times. What's your point? 05 meters cubed times, one out tio tools per kilogram k times our delta teams 1.8 degrees Celsius and that gives us Q is equal. Tio 118 Jules. So if the heat produced by the hamster So now we move on to Part B, which says, which asks us to find how many grams of seed the hamster needs per hour. So for part B, normally, in order to find how much energy something produces, we would do the mass times, the energy times, the efficiency, which in this case is your point is 10% 0.1. And that would equal R Q eve for efficiency here or energy. Excuse me, energy value. But in this case, we're actually solving for mass. We don't have mass is what we're looking for. So this becomes mass. Times 24 Jules per kilogram. Time 0.1 equals our 118 Jules. And then we rearrange that equation and we get Sol for mass. M equals 118. Jules over 0.1 times 24. Jules, her Graham. And that gives us a value of 49 grams. So is our hamsters. Wait at 50 grams. This doesn't really seem reasonable that he will able to eat his equal body weight every hour, but who knows?

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