The enthalpy of fusion of Ice is 6.02 kJ mol − 1 . The heat capacity of H 2 O is 4.18 Jg − 1 K − 1 . What is the smallest number of ice cubes at 0 ∘ C , each containing one mole of H 2 O , that ar

Question

The enthalpy of fusion of Ice is 6.02 kJ mol − 1 . The heat capacity of H 2 ​ O is 4.18 Jg − 1 K − 1 . What is the smallest number of ice cubes at 0 ∘ C , each containing one mole of H 2 ​ O , that are needed to cool 500 g of liquid water from 2 0 ∘ C to 0 ∘ C ?

TestHub · Accepted Answer

Heat released to cool 500 g water from 2 0 ∘ C to 0 ∘ C , Q = m s Δ T = 500 × 4.18 × 20 = 41 , 800 J = 41.8 kJ Number of moles of water (Ice) that will melt to absorb 41.8 k J = 6.02 41.8 ​ = 7 Number of cubes of ice that will melt = 7 .

Chemistry - Thermodynamics - I Question with Solution | TestHub

Doubts & Discussion

JEE Advanced (2028)