on a winter day the temperature drops from -15c too -25C overnight. if a pan sitting outside contains 0.50 kg of ice, how much heat is removed from the ice for the temperature change?
-15 to -25 = 10 degree C change.
q removed = mass ice x specific heat ice x 10
mass ice = 500 g
specific heat ice = 2.108 J/g*C
To calculate the amount of heat removed from the ice for the temperature change, you can use the formula:
Q = m * c * ΔT
Where:
Q = amount of heat transferred
m = mass of the ice (0.50 kg)
c = specific heat capacity of ice (2,090 J/kg·°C)
ΔT = change in temperature (final temperature - initial temperature)
First, calculate the change in temperature:
ΔT = -25°C - (-15°C)
ΔT = -25°C + 15°C
ΔT = -10°C
Now, substitute the values into the formula:
Q = 0.50 kg * 2,090 J/kg·°C * -10°C
Q = -5,225 J
Therefore, the amount of heat removed from the ice for the temperature change is -5,225 J. Note that the negative sign indicates that heat is being removed or lost from the ice.
To calculate the amount of heat removed from the ice for the temperature change, we need to use the equation:
Q = m * ΔT * C
Where:
Q represents the heat removed from the ice
m is the mass of the ice (0.50 kg)
ΔT is the change in temperature (-25°C - (-15°C) = -10°C)
C is the specific heat capacity of ice (2090 J/kg·°C)
Using the given values, we can calculate the heat removed from the ice:
Q = 0.50 kg * (-10°C) * 2090 J/kg·°C
Calculating this equation would yield the amount of heat removed from the ice during the temperature change.