Question

specific heat capacites

ice = 2100 JKG^-1K^-1

silver 234  JKG^-1K^-1

Latent heat of fusion of water = 3.35x10^5 JKg^-1

Answer 1

Given

mass of ice Mi = 1.0* 10^-2 kg

Latent heat of water = 3.35* 10^5 J Kg^-1

Specific of water = 2100 JKg^-1K^-1 , specific heat silver = 234JKg^-1K^-1

Temperature of silver before come into contact with ice = 55°C

Also given , ice is at melting temperature = 0°C

​​​​​​Applying energy conversion rule

Heat gained by ice for its phase change = Heat lost by mass of silver.

Since only 65% of ice got converted into water , so the ice will use only its latent heat of fusion and whole mass ( 65% water and 35% ice ) will remain at 0° C at equillibrium Temperature.

0.65 *Mi*L =Ms*Ss*(Ts-0) ( L= latent heat of fusion of water

( Ss= specific heat of silver)

0.65*1.0*10^-2 *3.35 * 10^5 = Ms*234 * (55-0)

Ms = 0.1692 Kg

Hence mass of silver required to melt ice 65% = 0.1692 Kg