Question

The body must burn calories to warm the water from 0°C to a body temperature of 37°C. Assume a dieter drinks 2.4 kg (i.e., 2.4 L) of ice water every day. The specific heat of water is 4.186 × 10³ J/(kg⋅°C) (1) How much energy, in joules, does the dieter’s body need to provide to warm the water over a full year. Take 365 days in a year (2) One pound of fat supplies 3400 kcal (Cal) to the body, where 1 kcal (Cal) = 4.186 × 10³ J. How much fat, in pounds, can the dieter hope to lose in a year by drinking ice water rather than water at body temperature?

Answer 1

In a single day, the energy needed to warm water, Q1 = m * C * T
Where m is the mass of water, C is the specific heat of water, and T is the change in temperature of the water.
Q1 = 2.4 * 4.186 * 103 * (37 - 0)
= 3.72 * 105 J
Energy needed for one year, Qw = 365 * Q1
= 365 * 3.72 * 105
= 1.36 * 108 J

Energy supplied by one pound fat, Qf = 3400 kcal
= 3400 * 4.186 * 103 J
= 0.142 * 108 J

Energy needed to warm water for one year = n * energy supplied by the fat
Qw = n * Qf
n = Qw/Qf
= (1.36 * 108) / (0.142 * 108)
= 9.53 pounds