Question

Physical therapists know as you soak tired muscles in a hot tub, the water will cool down as you heat up. If a 67.9 kg person at 37.1 oC immerses in 50.2 kg of water at 40.5o C, the equilibrium temperature is 38.7 oC.
What is the specific heat of the person?
Use 4.186 kJ/kgoC for the specific heat of water.

Answer 1

Answer:

c = 5032.8J/kg°C

Explanation:

the transfer of heat is given by

$Q_1=-Q_2\\\\Q=cm(T_f-T_i)$

c is the specific heat, m the mass and T the temperatures at equilibrium an at the beginning.

By replacing we obtain

$c_1m_1(T_f-T_{1i})=-c_2m_2(T_f-T_{2i})\\\\c_2=-\frac{c_1m_1(T_f-T_{1i})}{m_2(T_f-T_{2i})}=-\frac{(67.9kg)(4186J/kg\°C)(38.7\°C-37.1\°C)}{(50.2kg)(38.7\°C-40.5\°C)}\\\\c_2=5032.8\frac{J}{kg\°C}$

hope this helps!!