contestada

3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C, in an insulated container. The final temperature of the metal–liquid mixture at equilibrium is 28.3°C. What is the specific heat of the liquid? Neglect the heat capacity of the container.

Respuesta :

sebassandin

Answer:

[tex]Cp_{liquid}=2.54\frac{J}{g\°C}[/tex]

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

[tex]Q_{Ag}=-Q_{liquid}[/tex]

That in terms of the heat capacities, masses and temperature changes turns out:

[tex]m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})[/tex]

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

[tex]Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}[/tex]

Best regards.

Otras preguntas