Question

When resting, a person has a metabolic rate of about 3.0X10^5 joules per hour. The person is submered neck deep into a tub containing 1.2X10^3 kg of water at 21.00 degrees C. If the heat from the person only goes into the water, find the water temperature after half an hour.
Given : Specific heat of water at 15 degrees C. 4186 j/kgc

Answer 1

Answer:

21.02986 °C

Explanation:

c = Specific heat of water at 15 °C = 4186 J/Kg°C

$\Delta T$ = Change in temperature

$T_1$ = Initial temperature = 21 °C

$T_2$ = Final temperature

m = Mass of person = $1.2\times 10^3\ kg$

t = Time taken = 30 minutes

Heat is given by

$Qt=mc\Delta T\\\Rightarrow Qt=mc(T_2-T_1)\\\Rightarrow \frac{3\times 10^5}{60}\times 30=1.2\times 10^3\times 4186(T_2-21)\\\Rightarrow T_2-21=\frac{3\times 10^5\times 30}{60\times 1.2\times 10^3\times 4186}\\\Rightarrow T_2-21=0.02986\\\Rightarrow T_2=0.02986+21\\\Rightarrow T_2=21.02986\ ^{\circ}C$

The water temperature after half an hour is 21.02986 °C