how many kilojoules of heat must be transferred to a 480 g aluminum pizza pan to raise its temperature from 22 degree c to 234 degree c? The specific heat of aluminum in this temperature range is 0.97 J/g degree C.
I hate science
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
hope this helps~ :)
To calculate the amount of heat transferred, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred (in kilojoules, kJ)
m is the mass of the aluminum pizza pan (in grams, g)
c is the specific heat of aluminum (in joules/gram-degree Celsius, J/g°C)
ΔT is the change in temperature (in degree Celsius, °C)
Given:
m = 480 g
c = 0.97 J/g°C
ΔT = 234°C - 22°C = 212°C
Substituting the given values into the formula:
Q = 480 g * 0.97 J/g°C * 212°C
Now, let's perform the calculation:
Q = 98016 J
To convert J (Joules) to kJ (Kilojoules), divide the value by 1000:
Q = 98.016 kJ
Therefore, approximately 98.016 kilojoules of heat must be transferred to the 480 g aluminum pizza pan to raise its temperature from 22°C to 234°C.