Question

A 39 g block of ice is cooled to −67◦C. It is added to 518 g of water in an 64 g copper calorimeter at a temperature of 28◦C.

Find the final temperature. The specific heat of copper is 387 J/kg ·◦C and of ice is 2090 J/kg ·◦C . The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is4186J/kg·◦C.

Answer in units of ◦C.

Answer 1

Given,

mass of the ice, m_ice = 39 g = 0.039 kg

Initial temperature of ice, T₁ = -67 C

mass of the water, m_water = 518 g = 0.518 kg

mass of the copper, m_copper = 80 g = 0.08 kg

Let the final temperature be T_f

Specific heat of copper, c_copper = 387 J/kg.C

Specific heat of ice, c_ice = 2090 J/kg.C

Specific heat of water, c_water = 4186 J/kg.C

Latent heat of fusion, L = 3.33 * 10⁵ J/kg.C

Now,

Heat needed to raise the temperature of ice from -67 to 0 degrees

    + latent heat of fusion needed to convert ice into water

        + Heat needed to raise the temperature of water from 0 to T_f degrees

                                                  = Heat released when temperature of water decreases from 28 to T_f

                                                               + Heat released when temperature of copper decreases from 28 to T_f

=> m_ice*c_ice*(0-(-67)) + m_ice*L + m_ice*c_water*(T_f - 0) = m_water*c_water*(28 - T_f) + m_copper*c_copper*(28 - T_f)

=> 0.039*2090*67 + 0.039*3.33*10⁵ + 0.039*4186*T_f = 0.518*4186*(28 - T_f) + 0.08*387*(28 - T_f)

=> 5461.2 + 12987 + 163.3*T_f = 60713.7 - 2168.4*T_f + 866.9 - 30.96*T_f

=> 163.3*T_f + 2168.4*T_f + 30.96*T_f = 60713.7 + 866.9 - 5461.2 - 12987

=> 2362.7*T_f = 43132.4

=> T_f = (43132.4) / (2362.7)

          = 18.3 C

thus final temperature is 18.3 Celsius.