Question

A 100g sample of hot copper is placed in a coffee cup calorimeter containing 100 grams of water at room temperature. After some time the temperature of the water and the copper become a constant at 50˚C. Calculate the initial temperature of the copper piece. The specific heat of copper is 0.385 J/g˚C. The specific heat of liquid water is 4.184 J/g˚C.

Answer 1

Answer:

$356^{\circ}C$

Explanation:

When the hot copper and the liquid water reaches equilibrium, they have the same temperature; when this happens, the amount of heat released by the copper is equal to the amount of heat absorbed by the water:

$-Q_c = Q_w\\-m_c C_c (T_f-T_c) = m_w C_w (T_f-T_w)$

where

$m_c = 100 g$ is the mass of the copper

C_c=0.385 J/g˚C is the specific heat of copper

T_f=50˚C is the final temperature of both substances

$m_w = 100 g$ is the mass of the water

C_w=4.184 J/g˚C is the specific heat of the water

T_c is the initial temperature of the copper

T_w=20˚C is the initial temperature of the water (room temperature)

Solving for T_c, we find:

$T_c = T_f + \frac{m_w C_w (T_f-T_w)}{m_c C_c}=50^{\circ} +\frac{(100 g)(4.184 J/gC) (50C-20C)}{(100 g)(0.385 J/gC)}=356^{\circ} C$