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?

Question

Noah Aldrich · Accepted Answer

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,

#DeltaH_("fus") approx (6.0kJ)/(mol)#

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

#9.0g * (mol)/(18g) * (6.0kJ)/(mol) approx 3.0kJ#

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!

Caleb Adams · Answer

undefined The specific heat capacity of the metal can be calculated using the formula:

$ q_{\text{metal}} = -q_{\text{ice}} $

where $ q_{\text{metal}} $ is the heat gained by the metal, and $ q_{\text{ice}} $ is the heat lost by the ice.

The heat gained by the metal ($ q_{\text{metal}} $) can be calculated using the formula:

$ q_{\text{metal}} = mc\Delta T $

where:
- $ m $ is the mass of the metal (in grams)
- $ c $ is the specific heat capacity of the metal (in J/g°C)
- $ \Delta T $ is the change in temperature of the metal (final temperature - initial temperature)

The heat lost by the ice ($ q_{\text{ice}} $) can be calculated using the formula for heat of fusion:

$ q_{\text{ice}} = m_{\text{ice}} \times \Delta H_{\text{fusion}} $

where:
- $ m_{\text{ice}} $ is the mass of the ice that melted (in grams)
- $ \Delta H_{\text{fusion}} $ is the heat of fusion of ice (in J/g)

Once you have both $ q_{\text{metal}} $ and $ q_{\text{ice}} $, set them equal to each other and solve for the specific heat capacity of the metal ($ c $).