Question

how much heat is required to change 10g ice at 0c to steam at 100c? latent heat of fusion and vapour for H2O are 80cal/g and 540cal/g respectively. specific heat of water is 1 cal/g

Answer 1

Q=mL+mcΔt+mL'

=10*80+10*1*(100-0)+10*540

=800+1000+5400

=7200cal.=7.2kcal

L=heat of fusion of ice

L'=heat of vapourisation of water

Answer 2

Answer:

Total heat is required to change 10 g ice at 0°C to steam at 100°C is 7200 cal.

Explanation:

Latent heat of fusion of ice= $\Delta H_{fus}=80cal/g$

Heat required to melt 10 g of ice = Q

$Q=m\times \Delta H_{fus}$

$=10 g\times 80cal/g=800 cal$

Heat require to raise the temperature of water from 0°C to 100°C = Q'

Specific heat of water =c=  1 cal/g°C

ΔT =  100°C -  0°C  = 100°C

$Q'=mc\times \Delta T$

$=10 g\times 1 cal/g^oC\times 100^oC=1000 cal$

Latent heat of vaporization of water= $\Delta H_{vap}=540cal/g$

Heat required to vaporize 10 g of water = Q''

$Q''=m\times \Delta H_{vap}$

$=10 g\times 540 cal/g=5400 cal$

Total heat is required to change 10 g ice at 0°C to steam at 100°C.

$=Q+Q'+Q''=800 cal+1000 cal+5400 cal = 7200 cal$