Question

9. A mass of 35.05 g of H20 (s) at 273 K is dropped into 185 g of H20 l) at 310 Kin an insulated contacting at 1 bar of pressure. Calculate the temperature of the system once equilibrium has been reached. Assume Cp.m for H20 (1) is constant at its values for 298K throughout the temperature range. Given: Cpm(ice) = 75.3 JK-mol-1

Answer 1

We know that, The heat lost by the liquid water = heat gained by ice (which is used to melt the ice, and then to rise the temperature of mixture to final equilibrium temperature, T)

Heat of fusion of ice,  ΔH = 334 J/g

Specific heat capacity of ice, C_i = 2.108 J/g.K

Hence, we can write the equation as:

H_ice = H_water

(m_iΔH) + (m_iC_i (T-273)) = m_lC_l (310-T)

Substituting given values:

(35.05 x 334) + ( 35.05 x 2.108​​​​​​​ x (T-273)) = 185 x 4.18 x (310-T)

11706.7 + 73.88 (T-273) = 773.3 (310-T)

We need to solve for T:

T = (11706.7 - 73.88 x 273 - 773.3 x 310) / (73.88 - 773.3)

∴ T = 293 K

• That is, Final equilibrium temperature of the mixture will be = 293 K