When 1 gram of liquid water at 0 degrees Celsius freezes to form ice, how many total calories of heat are lost by the water and how do you know?
a. 1
b. 0.5
c. 80
d. 540
c. (80 calories per gram), because that is what it is. It is a value obtained experimentally.
To determine the total calories of heat lost when 1 gram of liquid water at 0 degrees Celsius freezes to form ice, we can use the specific heat capacity and enthalpy of fusion values for water.
The specific heat capacity of water is 1 calorie/gram-degree Celsius (cal/g°C). This indicates that it takes 1 calorie of energy to increase the temperature of 1 gram of water by 1 degree Celsius.
The enthalpy of fusion of water is 80 calories/gram (cal/g). This value represents the amount of heat energy released or absorbed when 1 gram of a substance changes state between solid and liquid at a constant temperature.
When liquid water freezes to form ice, it undergoes a change in state from liquid to solid. This means that the water releases heat energy equal to the enthalpy of fusion.
Therefore, the correct answer is:
c. 80 calories
To determine the total calories of heat lost by the water when it freezes, we need to calculate the heat of fusion for water.
The heat of fusion is the amount of heat required to convert a substance from a solid to a liquid or vice versa at a constant temperature. For water, the heat of fusion is 80 calories/gram.
Given that we have 1 gram of liquid water and it freezes to form ice at 0 degrees Celsius, we can calculate the total calories of heat lost as follows:
Total calories of heat lost = Heat of fusion x Mass of water
= 80 calories/gram x 1 gram
= 80 calories
Therefore, the correct answer is c. 80.