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How many joules of heat must be transferred to a 410-g aluminum pizza pan to raise its temperature from 32oC to 232oC? The specific heat of aluminum in this temperature range is 0.96 J/g x oC.

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Answer:

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

Step-by-step explanation:

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