What is the final temperature, in oC, after a 26.5 g piece of ice, at 0oC, is placed in a styrofoam cup with 122 g of water initially at 73.8oC

the specific heat of water = 4.184 JK-1g-1
The heat of fusion of ice = 333 Jg-1

heat to melt ice + heat to raise T of melted ice from zero C to the final T + heat lost by warm water = 0

1st term:
mass ice x deltaH fusion.

2nd term:
mass melted ice x specific heat water x (Tfinal-Tinitial)

3rd term:
mass hot water x specific heat water x (Tfinal-Tinitial)

Add all of them, set to zero, and solve for the only unknown; i.e., Tfinal.

To find the final temperature, we need to use the principle of conservation of energy. The heat lost by the water and ice system is equal to the heat gained by the surrounding environment (styrofoam cup).

First, let's calculate the heat lost by the water:

Q_water = m_water * c_water * ΔT_water

where:
- Q_water is the heat lost by the water
- m_water is the mass of water = 122 g
- c_water is the specific heat of water = 4.184 J/(g°C)
- ΔT_water is the change in temperature of water = final temperature - initial temperature = final temperature - 73.8°C

Next, let's calculate the heat gained by the ice during the process of reaching the final temperature:

Q_ice = m_ice * L_fusion + m_ice * c_ice * ΔT_ice

where:
- Q_ice is the heat gained by the ice
- m_ice is the mass of ice = 26.5 g
- L_fusion is the heat of fusion of ice = 333 J/g
- c_ice is the specific heat of ice (which is negligible compared to water's specific heat) = 2.09 J/(g°C)
- ΔT_ice is the change in temperature of ice = final temperature - 0°C

Since the heat lost by the water is equal to the heat gained by the ice, we can set up an equation:

Q_water = Q_ice

m_water * c_water * ΔT_water = m_ice * L_fusion + m_ice * c_ice * ΔT_ice

Substituting the given values:

122 g * 4.184 J/(g°C) * (final temperature - 73.8°C) = 26.5 g * 333 J/g + 26.5 g * 2.09 J/(g°C) * (final temperature - 0°C)

Now we can solve this equation to find the final temperature.