A 200-g metal container, insulated on the outside, holds 100 g of water in thermal equilibrium at 22.00°C. A 21-g ice cube, at the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.00°C. Assume there is no heat exchange with the surroundings. For water, the specific heat is 4190 J/kg · K and the heat of fusion is 3.34 × 10⁵ J/kg. The specific heat for the metal is closest to

Question

Quizplus · Accepted Answer

First, we can calculate the amount of heat released by the ice as it melts:
$q_{ice} = m_{ice} \cdot L_f = 21	ext{ g} \cdot 334	ext{ J/g} = 7026	ext{ J}$
where $L_f$ is the heat of fusion.

This heat is absorbed by the water as the ice melts and by the water and container as the resulting water cools from 22.00°C to 15.00°C:
$q_{water+container} = (m_{water} c_{water} + m_{container} c_{container}) \Delta T$
where $\Delta T = 22.00	ext{°C} - 15.00	ext{°C} = 7.00	ext{°C}$.

We know the mass of the water and the ice, but not the mass or specific heat of the container. However, we can use the fact that the container is insulated to assume that there is no heat exchange with the surroundings. This means that the total amount of heat in the system (water, ice, and container) is conserved:

$q_{ice} = -q_{water+container}$

Plugging in the values we know and solving for $c_{container}$, we get:

$c_{container} = \frac{q_{water+container}}{m_{container} \Delta T} = \frac{-q_{ice}}{m_{container} \Delta T} = \frac{-7026	ext{ J}}{200	ext{ g} \cdot 7.00	ext{°C}} \approx 3850	ext{ J/kg} \cdot 	ext{K}$

Therefore, the closest choice for the specific heat of the container is (A) 3850 J/kg ∙ K.

A 200-G Metal Container, Insulated on the Outside, Holds 100