
William W. answered 04/09/23
Experienced Tutor and Retired Engineer
The aluminum and water gained heat:
Q = (mCpΔT)A + (mCpΔT)W
Q = (100)(0.890)(20 - 10) + (250)(4.186)(20 - 10) = 11355 joules
The metal lost the same amount of heat:
Q = (mCpΔT)1 + (mCpΔT)2
-11355 = (75)(0.385)(20 - 60) + (70)(x)(20 - 100)
-11355 = -1155 - 5600x
x = 1.82 j/g/°C
You are correct