A52.0−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

it is a good conductor

explanation: