Question

you take 295.5 g of a solid at 30.0 c and let it melt in 425 g of water. the water temperature decreases from 85.1 c to 30.0 c. calculate the heat of fusion of this solid

Answer 1

Answer:

$\Delta _{fus}H=332 \frac{J}{g}$

Explanation:

Hello!

In this case, since the heat of fusion of a solid substance stands for the energy required to melt it, which is a phase transition from solid to liquid, we can see that the heat lost by water is that gained by the solid, so we can write:

$Q_{solid}=-Q_{w}$

Thus, by using the water data and its specific heat (4.184), we obtain:

$Q_{solid}=-m_{w}C_{w}(T_f-T_i)\\\\Q_{solid}=-425g*4.184\frac{J}{g\°C}*(30.0-85.1)\°C\\\\ Q_{solid}=97,978.82J=98.0kJ$

Next, since the heat of fusion of a substance is usually represented in terms of energy per amount of substance, we use the mass of solid to obtain:

$\Delta _{fus}H=\frac{98.0kJ}{295.5g}\\\\ \Delta _{fus}H=0.332\frac{kJ}{g}=332 \frac{J}{g}$

Best regards!