Question

4 kg of ice at -20 degree Celsius is mixed with 10kg of water at 20 degree Celsius in an insulating vessel having a neglifible heat capacity. calculate the final mass of water remaining in the container. Given specific heat capacity of water and ice is 4.184 kJ/K/kg and 2.092kJ/K/kg. molar enthalpy of fusion of water of ice is 334.7kJ/kg

Answer 1

Dear student!
At first, the ice will melt to liquid water, so to calculate total amount of liquid we take the following formula as:
m_iC_i ΔT + m_lΔH_fus - m_wC_w ΔT = 0 (insulated system)
Here, m_iC_i and m_iC_i are the masses and specific heats of ice and water respectively and m_l is the mass of total liquid and
ΔT = change in temperature (final - initial)
For ice ΔT = 273 (0 ^oC)- 253 = 20K, for water ΔT = 20 K (293-273K)
Hence, putting the values we get,
4 x 2.092 x 20 + m_l x334.7 - 10 x 4.184 x 20 = 0
167.36 + 334.7 m_l - 836.8 = 0
or, 334.7 m_l =838.6 - 167.36 = 669.44
Hence, m _l = 669.44/ 334.7 = 2.00Kg
Hence, total mass of liquid water in the container, m = 10 Kg + 2Kg = 12Kg.