How much heat must you remove if you want to lower the temperature of 350 g of steel (c = 0.11 cal/gºC) from 18.0 ºC to 12.0 ºC? Express this answer in both calories and Joules.

Answer 1

q = mC∆T

q = heat = ?

m = mass = 350 g

C = specific heat = 0.11 cal/gº

∆T = change in temperature = -6º

q = (350 g)(0.11 cal/gº)(-6º) = -231 cal

Since 1 cal = 4.184 joules (J), we have the following conversion:

-231 cal x 4.184 J/cal = -967 J

So the answer would be 231 cal or 967 joules must be removed from the steel (the negative sign indicates that heat is being removed)

Answer 2