Question

101. Du

Answer 1

Answer:

$0.972 J/g^{\circ}C$

Explanation:

When the sample of hot metal is placed inside the water, the metal gives off heat to the water: the temperature of the metal decreases while the temperature of the water increases, until the two reach an equilibrium temperature.

The heat given off by the metal is equal (in magnitude) to the heat absorbed by the water, so we can write:

$Q_m=Q_w\\m_m C_m (T_m-T_e) = m_w C_w (T_e-T_w)$

where:

$m_m=26.0 g$ is the mass of the metal

$C_m$ is the specific heat of the metal

$T_m=82.25^{\circ}C$ is the initial temperature of the metal

$T_e=28.34^{\circ}C$ is the equilibrium temperature

$m_w=75.0 g$ is the mass of the water

$C_w=4.186 J/g^{\circ}C$ is the specific heat of water

$T_w=24.00^{\circ}C$ is the initial temperature of the water

And solving for $C_m$, we find the specific heat of the metal:

$C_m=\frac{m_w C_w (T_e-T_w)}{m_m(T_m-T_e)}=\frac{(75.0)(4.186)(28.34-24.00)}{(26.0)(82.25-28.34)}=0.972 J/g^{\circ}C$