A piece of ice of mass `50 g` exists at a temperature of `-20^(@)C`. Determine the total heart required to covert it completely to steam at `100^(@)C`. (Specific heat capacity of ice `= 0.5 cal//g-^(@)C`, specific latent heat of fusion for ice `= 80 cal//g` and specific latent heat of vaporization for water `= 540 cal//g`).

Options

A. `26500 cal`
B. `12400 cal`
C. `36500 cal`
D. `46500 cal`

Answer

C

Explanation

For changing ice at `-20^(@)C` to steam at `100^(@)C` the complete process is achived in four different stages :
(i) The rise in temperature of ice from `-20^(@)C` to `0^(@)C` (i.e., melting point of ice).
Heat required for this process
`DeltaQ_(1) = (50g)(0.5 cal//g-^(@)C) (0+20)^(@)C = 500cal`
(ii) The melting of ice at `0^(@)`
Heat required for this process
`Delta Q_(2) = (50g)(80cal//g)=4000cal`
(iii) The rise in temperature of melted ice (i.e. water) from `0^(@)` to `1000^(@)C` (i.e., B.P of water)
Heat required for this process
`Delta Q_(3) = (50g)(1 cal//-^(@)C)(100^(@)C)=5000cal`
(iv) The vaporisation of water at `100^(@)C`
Heat required for this process
`Delta Q_(4) = (50g)(540 cal//g)=27000cal`
`:.` Total heat required `Delta Q = Delta Q_(1)+Delta Q_(2)+Delta Q_(3)+Delta Q_(4)`
`=36500 cal`.