Question

The specific heat of a human is approximately 3.47 J/°C. Use this information to answer the following questions.

(a) If a 1601 lb man eats a candy bar containing 287 cal, how much will his body temperature increase if all of the calories from the candy bar are converted into heat energy?

(b) If a 160 lb man eats a roll of candy containing 41.9 cal, how much will his body temperature increase if all of the calories from the candy are converted into heat energy?

Answer 1

The body temperature of a 1601 lb man will increase by approximately 0.476°C after eating a 287 cal candy bar, and the body temperature of a 160 lb man will increase by approximately 0.696°C after eating a 41.9 cal candy roll, assuming all the calories are converted to heat energy.

Step-by-step explanation:

To solve for the temperature increase caused by consuming calories, we use the equation:

Q = mcΔT,

where Q is the heat energy (in joules), m is the mass (in kilograms), c is the specific heat capacity, and ΔT is the change in temperature (in Celsius).

(a) For the 1601 lb man consuming a 287 cal candy bar, first, we need to convert pounds to kilograms (1 lb = 0.453592 kg) and calories to joules (1 cal = 4.184 J).

Mass of man = 1601 lb * 0.453592 = 726 kg (approx.),

Calories to joules = 287 cal * 4.184 J/cal = 1201 J (approx.)

Now we can calculate the temperature increase:

1201 J = 726 kg * 3.47 J/kg/°C * ΔT,

ΔT = 1201 J / (726 kg * 3.47 J/kg/°C) = 0.476 °C (approx.)

(b) For the 160 lb man consuming a 41.9 cal candy, we have:

Mass of man = 160 lb * 0.453592 = 72.6 kg (approx.),

Calories to joules = 41.9 cal * 4.184 J/cal = 175 J (approx.)

ΔT = 175 J / (72.6 kg * 3.47 J/kg/°C) = 0.696 °C (approx.)