On a winter day the temperature drops from –5°C to –15°C overnight. If a pan sitting outside contains 0.49 kg of
ice, how much heat is removed from the ice for this temperature change?
Heat out = Specific heat of ice * mass of ice * change in temp
To find out how much heat is removed from the ice for this temperature change, we can use the equation:
Q = mcΔT
where:
Q = heat removed from the ice (in joules)
m = mass of the ice (in kilograms)
c = specific heat capacity of ice (2090 J/kg°C)
ΔT = change in temperature (in °C)
First, let's calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = (-15°C) - (-5°C)
ΔT = -15°C + 5°C
ΔT = -10°C
Now, we can substitute the known values into the equation:
Q = (0.49 kg) * (2090 J/kg°C) * (-10°C)
Calculating this equation will give us the amount of heat removed from the ice.