On a winter day the temperature drops from –15°C to –25°C overnight. If a pan sitting outside contains 0.40 kg of ice, how much heat is removed from the ice for this temperature change?
To determine the amount of heat removed from the ice for this temperature change, we need to use the equation:
Q = mcΔT
Where:
Q is the amount of heat (in joules),
m is the mass of the object (in kilograms),
c is the specific heat capacity of the material (in joules per kilogram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
In this case, the substance is ice and we need to find the amount of heat removed. The specific heat capacity of ice is approximately 2,090 J/(kg°C), which means that it takes 2,090 Joules to change the temperature of 1 kilogram of ice by 1 degree Celsius.
Given:
Mass of ice (m) = 0.40 kg
Change in temperature (ΔT) = -25°C - (-15°C) = -25°C + 15°C = -10°C
Specific heat capacity of ice (c) = 2,090 J/(kg°C)
Now we can substitute the values into the equation:
Q = (0.40 kg) × (2,090 J/(kg°C)) × (-10°C)
Q = -8,360 J
The negative sign indicates that heat is being removed from the ice. Therefore, approximately 8,360 Joules of heat is removed from the ice for this temperature change.