# A 25 g piece of an unknown metal alloy at 150°C is dropped into an insulated container with 200 g of ice. How do you calculate the specific heat capacity of the metal, given that 9.0 g of ice melted?

Dec 30, 2017

You never gave us the temperature of the ice, so I will assume it's approximately 0°C.

Moreover, we'll need the enthalpy of fusion for water,

$\Delta {H}_{\text{fus}} \approx \frac{6.0 k J}{m o l}$

Let's figure out how much heat was used to melt that ice,

$9.0 g \cdot \frac{m o l}{18 g} \cdot \frac{6.0 k J}{m o l} \approx 3.0 k J$

Moreover, let's assume that the metal only released that heat and equilibrated with the ice,

3000J = 25g * C_s * 150°C

therefore C_s approx (0.8J)/(g*°C)

This is fairly reasonable, most metals' specific heats are between 0 and 1. I'm open to feedback if I made a mistake!