An ice cube of mass 100 g and at 0C is dropped into a Styrofoam cup containing 200 g of water at 25C. The heat of fusion of ice is 80 cal/g and the specific heat capacity of water is 1.0 cal/g C°. Assuming the cup doesn't exchange any heat, the final temperature of the system will be which of the following?
To find the final temperature of the system, we need to use the principle of conservation of energy. The heat lost by the water as it cools down will be equal to the heat gained by the ice as it warms up and melts.
First, let's calculate the heat lost by the water. We can use the formula:
Q_water = mass_water * specific heat capacity_water * change in temperature
Given:
mass_water = 200 g
specific heat capacity_water = 1.0 cal/g °C
change in temperature = (final temperature - initial temperature)
Since the cup doesn't exchange any heat, the final temperature of the system will be the same as the initial temperature of the ice, which is 0°C.
Q_water = 200 g * 1.0 cal/g °C * (0°C - 25°C)
= 200 g * 1.0 cal/g °C * (-25°C)
= -5000 cal
Next, let's calculate the heat gained by the ice as it warms up and melts. We can use the formula:
Q_ice = mass_ice * specific heat capacity_ice * change in temperature + mass_ice * heat of fusion
Given:
mass_ice = 100 g
specific heat capacity_ice = 1.0 cal/g °C
change in temperature = (final temperature - initial temperature)
heat of fusion = 80 cal/g
Q_ice = 100 g * 1.0 cal/g °C * (0°C - 25°C) + 100 g * 80 cal/g
= 100 g * 1.0 cal/g °C * (-25°C) + 100 g * 80 cal/g
= -2500 cal + 8000 cal
= 5500 cal
According to the principle of conservation of energy, the heat lost by the water is equal to the heat gained by the ice:
Q_water = Q_ice
-5000 cal = 5500 cal
Since the equation is not balanced, we made an error in our calculations or assumptions. Let's review our solution to find the mistake.