Question

In an experiment, 22.0 g of metal was heated to 98.0°C and then quickly transferred to 150.0 g of water in a calorimeter. The initial temperature of the water was 27.0°C, and the final temperature after the addition of the metal was 32.5°C. Assume the calorimeter behaves ideally and does not absorb or release heat.

Answer 1

q_{lost, metal} = q_{gained, water}

q = m C_p Δt   ....{ m = mass, Cp = specific heat capacity, Δt = Tfinal - T initial }   

(mass of metal) (Δt) (C_{p, metal}) = (mass of water) (Δt) (C_{p, water})

molar heat capacity of metal = ? J/mol ^oC

heat capacity of water = 4.18 J/g ^oC

(22.0 g) x (Cp) x (98.0 ^oC - 32.5 ^oC) = (150.0 g ) x (4.18 J/g^oC) x (32.5 ^oC - 27.0^oC)

1441 (Cp) = 413448.5

heat capacity of metal = 2.34 J/g ^oC

heat gained by water q = m Cp ΔT

q = 150 g x 4.18 J/g oC x (32.5 - 27.0) ^oC

q = 3448.5 J = 3.4485 KJ