Question

How much heat (in kJ) is required to warm 11.0 g of ice, initially at -11.0 ∘C, to steam at 110.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

Answer 1

While converting ice at- 11 deg.c to water vapor at 110 deg.c ( Assuming a pressure of 1 atm). Sensible heat has to be supplied for converting ice from -11 deg.c to 0 deg.c, then latent heat has to be supplied for converting ice at 0 deg.c to liquid at 0 deg.c, then sesible heat from 0 to 100 deg.c, latent heat at 100 deg.c and finally sensible heat of vapor from 100 deg,c to 110 deg.c)

1. Sensible heat of ice = Mass *specific heat* temperature differecne= 11*2.09*(0-(-11)= 252.89 joules
2. Latent heat of fusion has to be supplied to convert at o deg.c to convert ice from solid to liquid. Latent heat of fusion of ice = 335j/g. Hence heat of fusion = 335*11 = 3685 J
3. Sensible hea of water ( from 0 deg.c to 100 deg.c)= 11*4.18* 100 ( since specific heat of liquid water =4.18 j/g.deg.c )= 4598 Joules
4. Latent heat of water is 2257 j/g and hence heat of vaporization= 2257*11=24827 joules
5. Sensible heat of vapor from (100 deg.c to 110 deg.c)= 11*2.01*(110-10)= 221.1 Joules

Total heatto be supplied = sum of 1+2+3+4+5= 252.89+3685+4598+24827+221.1 joules =33583.99 joules= 33.584 Kj