A 52.0-g sample of an unknown metal at 99°C was placed in a constant-pressure calorimeter containing 75.0 g of water at 24.0°C. The final temperature of the system was found to be 28.4°C. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 14.4 J/°C.)

The specific heat of the metal is 0.39251 J/g°C

Explanation:

The heat transfered from the metal to the calorimeter and water (Qwater + Qcal), must be equal to Qmetal (but different in direction).

Qmetal = - (Qwater + Qcal)

For Q we use the following formule Q = m * Cp*ΔT

with m = the mass in grams

with Cp = the heat capacity in J / g°C

with ΔT = the change in temperature = Final temperature T2 - initial temperature T1

Qmetal = m (Metal) * Cp (metal) * ΔT (metal)

with mass = 52 grams

with Cp = To be determined

with ΔT = 28.4 - 99 = - 70.6 °C

QCal = Cp (cal) * ΔT (water)

Qcal = 14.4*4.184 = 60.2496

Qwater = m (water) * Cp (water) * ΔT (water)

with mass = 75 g

with Cp (water) = 4.184

with ΔT (water) = 28.4 - 24 = 4.4

Qwater = 1380.72

Q (metal) = - (Qwater + Qcal)

m (Metal) * Cp (metal) * ΔT (metal) = - 60.2496 - 1380.72

Cp (metal) = - 1440.9696 / (52 * - 70.6)

Cp (metal) = - 1440.9696 / - 3671.2

Cp (metal) = 0.39251 J/g°C

The specific heat of the metal is 0.39251 J/g°C