Question

What is the amount of heat absorbed by 25 g of ice at -10°C when it is converted to water at 20°C? Given the enthalpy of fusion of ice is 6.01 kJ/mol and the specific heat capacity of water is 4.18 J/g°C.

Answer 1

To calculate the amount of heat absorbed by the ice, we need to consider three steps:1. Heating the ice from -10C to 0C2. Melting the ice at 0C3. Heating the water from 0C to 20CStep 1: Heating the ice from -10C to 0CQ1 = mass x specific heat capacity of ice x temperature changeSpecific heat capacity of ice is approximately 2.1 J/gC.Q1 = 25 g x 2.1 J/gC x  0 -  -10  C = 25 x 2.1 x 10 = 525 JStep 2: Melting the ice at 0CFirst, we need to find the number of moles of ice:Molar mass of water  H2O  = 18.015 g/molmoles = mass / molar mass = 25 g / 18.015 g/mol  1.387 molQ2 = moles x enthalpy of fusionQ2 = 1.387 mol x 6.01 kJ/mol = 8.34 kJ = 8340 J  converted to Joules Step 3: Heating the water from 0C to 20CQ3 = mass x specific heat capacity of water x temperature changeQ3 = 25 g x 4.18 J/gC x  20 - 0 C = 25 x 4.18 x 20 = 2090 JNow, we sum up the heat absorbed in all three steps:Total heat absorbed  Q  = Q1 + Q2 + Q3 = 525 J + 8340 J + 2090 J = 10955 JSo, the amount of heat absorbed by 25 g of ice at -10C when it is converted to water at 20C is 10,955 Joules.