Question

A 5.00 g piece of metal is heated to 100.0 C then placed in a beaker containing 20.0 g of water at 10.0 C. The temperature of the water rises to 15.0 C. Calculate the specific heat of the metal

Answer 1

Calculate the heat gained by the water first.

q = mCpΔT
m = 20.0 g
Cp = 4.186 J/g°C
ΔT = T(final) - T(initial) = 15.0°C - 10.0°C = 5.0°C

q = (20.0)(4.186)(5.0) = 419 J

This is equal to the heat lost by the metal, so calculate Cp for the metal, given:
q = -419 J (negative because heat was lost)
m = 5.00 g
ΔT = 15.0°C - 100.0°C = -85.0°C (negative because the temperature decreased)

q = mCpΔT —> Solve for Cp —> Cp = q/mΔT

Cp = -419 / (5.00 • -85.0) = 0.986 J/g°C