Question

How much energy (in kilojoules) is needed to heat 4.65 g of ice from -11.5 ∘C to 20.5 ∘C? The heat of fusion of water is 6.01kJ/mol, and the molar heat capacity is 36.6 J/(K⋅mol) for ice and 75.3 J/(K⋅mol) for liquid water.

Answer 1

For converting ice at -11.5 deg.c to 20.5 deg.c water, the following heat energy has to be supplied

1. heat energy for converting ice at -11. 5 deg.c to 0 deg.c ( sensible heat)= moles of water * specific heat of ice ( in j/k.mol) temperature difference= (4.65/18) moles* 36.6 j/K.nol* {0-(-11.50}=108.725 Joules= 0.1087 Kj

2. Latent heat needs to be supplied for converting ice at 0 deg.c to water at 0 deg.c and this amount is =Latent heat of ice* moles of water= 6.01( Kj/mol)*4.65/18= 1.5525 KJ

3. Heat energy required for converting water at 0 deg.c to water at 20.5 deg.c includes sensible heat and this is given b y= (4.65/18)*75.3* (20.5-0) joules = 0.399 KJ

Total heat to be suppled is sum of 1+2+3= 0.1087+1.5525+0.399=2.0062 Kj