Exemplo de Physics, 11.3 Phase Change and Latent Heat

Calculating Final Temperature from Phase Change: Cooling Soda with Ice Cubes

Problema e solução em inglês, como no livro original.

Ice cubes are used to chill a soda at 20 $°C$ and with a mass of $m_{s o d a} = 0.25 kg$ . The ice is at 0 $°C$ and the total mass of the ice cubes is 0.018 kg. Assume that the soda is kept in a foam container so that heat loss can be ignored, and that the soda has the same specific heat as water. Find the final temperature when all of the ice has melted.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
The ice cubes are at the melting temperature of 0 $°C$ . Heat is transferred from the soda to the ice for melting. Melting of ice occurs in two steps: first, the phase change occurs and solid (ice) transforms into liquid water at the melting temperature; then, the temperature of this water rises. Melting yields water at 0 $°C$ , so more heat is transferred from the soda to this water until they are the same temperature. Since the amount of heat leaving the soda is the same as the amount of heat transferred to the ice.

$Q_{i c e} = - Q_{s o d a}$
11.20

The heat transferred to the ice goes partly toward the phase change (melting), and partly toward raising the temperature after melting. Recall from the last section that the relationship between heat and temperature change is $Q = m c Δ T$ . For the ice, the temperature change is $T_{f} - 0 °C$ . The total heat transferred to the ice is therefore

$Q_{i c e} = m_{i c e} L_{f} + m_{i c e} c_{w} (T_{f} - 0 °C) .$
11.21

Since the soda doesn’t change phase, but only temperature, the heat given off by the soda is

$Q_{s o d a} = m_{s o d a} c_{w} (T_{f} - 20 °C) .$
11.22

Since $Q_{i c e} = - Q_{s o d a}$ ,

$m_{i c e} L_{f} + m_{i c e} c_{w} (T_{f} - 0 °C) = - m_{s o d a} c_{w} (T_{f} - 20 °C) .$
11.23

Bringing all terms involving $T_{f}$ to the left-hand-side of the equation, and all other terms to the right-hand-side, we can solve for $T_{f}$ .

$T_{f} = \frac{m_{s o d a} c_{w} (20 °C) - m_{i c e} L_{f}}{(m_{s o d a} + m_{i c e}) c_{w}}$
11.24

Substituting the known quantities
$T_{f} = \frac{(0.25 kg) (4186 J/kg \cdot °C) (20 °C) - (0.018 kg) (334,000 J/kg)}{(0 .25 kg + 0 .018 kg) (4186 K/kg \cdot °C)} = 13 °C$
11.25

Como foi?

Estratégia

Estratégia