A piece of ice of mass 50 g exists at a temperature of -20° C. Determine the total heat required to convert 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 vaporisation for water =540 cal / g )

Question

A piece of ice of mass 50 g exists at a temperature of -20° C. Determine the total heat required to convert 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 vaporisation for water =540 cal / g ) (A) 26500 cal (B) 12400 cal (C) 36500 cal (D) 46500 cal

Tardigrade Admin · Accepted Answer

For changing ice at -20° C to steam at 100° C the complete process is achieved 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 Δ Q1=(50 g )(0.5 cal / g - ° C )(0+20)° C =500 cal (ii) The melting of ice at 0° C Heat required for this process Δ Q2=(50 g )(80 cal / g )=4000 cal (iii) The rise in temperature of melted ice (i.e., water) from 0° C to 1000° C. (i.e., B.P. of water) Heat required for this process Δ Q3=(50 g )(1 cal / g - ° C )(100° C )=5000 cal (iv) The vaporisation of water at 100° C Heat required for this process Δ Q4=(50 g )(540 cal / g )=27000 cal ∴ Total heat required Δ Q=Δ Q1+Δ Q2+Δ Q3+Δ Q4 =36500 cal