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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.