Question

You drop an ice cube into an insulated container full of water and wait for the ice cube to completely melt. The ice cube initially has a mass of 80.0 g and a temperature of 0°C. The water (before the ice cube is added) has a mass of 660 g and an initial temperature of 20.0°C. What is the final temperature (in °C) of the mixture? (Assume no energy is lost to the walls of the container, or to the environment.)

A large room in a house holds 947 kg of dry air at 30.2°C. A woman opens a window briefly and a cool breeze brings in an additional 50.4 kg of dry air at 13.2°C. At what temperature (in degrees Celsius) will the two air masses come into thermal equilibrium, assuming they form a closed system? (The specific heat of dry air is

1006

J

kg · °C

, although that value will cancel out of the calorimetry equation.)

A rectangular glass window pane on a house has a width of 1.4 m, a height of 1.8 m, and a thickness of 0.46 cm. Find the energy transferred through the window by conduction in 12 hours on a day when the inside temperature of the house is 22°C and the outside temperature is 2.0°C. Take surface air layers into consideration.

Answer 1

given
m_ice = 80 g = 0.080 kg
m_water = 660 g = 0.66 kg

The specific heat of water (C) = 4186 J/(kg K)
Latent heat of water is L_f = 3.33*10^5 J/kg

(1)

Consider the final equilibrium temperature is T_f.

Apply the conservation of energy,

heat lost by water = heat gained by ice

m_water*C_water*(20 - Tf) = m_ice*Lf + m_ice*C_water*(Tf - 0 )

0.66*4186*(20 - Tf) = 0.080*3.33*10^5 + 0.080*4186*Tf

55255.2- 2762.76*Tf = 26640 + 334.88*Tf

Tf = 9.24 degrees celsius