M ~kg of water at t^ C is divided into two parts so that one part of mass m ~kg when converted into ice at 0^ C would release enough heat to vapourise the other part, then m M is equal to (Specific heat of water =1 cal g ⁻¹ ^ C ⁻¹ , Latent heat of fusion of ice =80 cal g ⁻¹ , Latent heat of steam =540 cal g ⁻¹ )

Step-by-step solution

Specific heat of water =1 cal g ^ -1 C ⁻¹ Latent heat of fusion of ice =80 cal g ⁻¹ Latent heat of steam =540 cal g ⁻¹ aligned & m 80+m 1 t=( M -m) 1 (100-t)+540( M -m) & m 80+m t=( M -m) 100-( M -m) t+540 M - m 540 & 80 m+m t= M 100-m 100- M t+m t+540 M -540 ~m & 80 m+100 m+540 m=640 M - M t & or, 720 m= M (640-t) & m M = 640-t 720 aligned