Question

Calculate the amount of heat absorbed when 20 grams of ice at -10°C is melted and then warmed to 30°C. The specific heat of ice is 2.09 J/g°C, the specific heat of water is 4.18 J/g°C and the enthalpy of fusion of ice is 6.01 kJ/mol.

Answer 1

To calculate the total amount of heat absorbed, 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 30CStep 1: Heating the ice from -10C to 0Cq1 = mass x specific heat of ice x Tq1 = 20 g x 2.09 J/gC x  0 -  -10  Cq1 = 20 g x 2.09 J/gC x 10Cq1 = 418 JStep 2: Melting the ice at 0CFirst, we need to find the moles of ice:Molar mass of water  H2O  = 18.015 g/molmoles of ice = mass / molar massmoles of ice = 20 g / 18.015 g/molmoles of ice = 1.11 molNow, we can calculate the heat absorbed during melting:q2 = moles x enthalpy of fusionq2 = 1.11 mol x 6.01 kJ/molq2 = 6.67 kJConvert kJ to J:q2 = 6.67 kJ x 1000 J/kJq2 = 6670 JStep 3: Heating the water from 0C to 30Cq3 = mass x specific heat of water x Tq3 = 20 g x 4.18 J/gC x  30 - 0 Cq3 = 20 g x 4.18 J/gC x 30Cq3 = 2512 JNow, we can find the total heat absorbed by adding the heat absorbed in each step:q_total = q1 + q2 + q3q_total = 418 J + 6670 J + 2512 Jq_total = 9600 JSo, the total amount of heat absorbed is 9600 J.