M grams of steam at 100^(@)C is mixed with 200g of ice at its melting point ina thermally insulated container. If it produces liquid water at 40^(@)C [heat of vaporization of water is 540cal/g and heat of fusion of ice is 80 cal/gl, the value of M is

M grams of steam at 100^(@)C is mixed with 200g of ice at its melting

1.	M grams of steam at 100^(@)C is mixed with 200g of ice at its melting point ina thermally insulated container. If it produces liquid water at 40^(@)C [heat of vaporization of water is 540cal/g and heat of fusion of ice is 80 cal/gl, the value of M is
Answer» 30 40 50 60 Solution :Suppose mass of ICE M= 200GM, mass of steam= m temperature of ice `T= 0^(@)C` temperature of steam `T= 100^(@)C` `ML_(f) + M(T. -T) C_(W) = mL_(v) + m(T.. - T.) C_(w)` `:. 200 XX 80 + 200 (40-0)1 = m xx 540 + m(100-40)(1)` `16000+ 8000= 540 m + 60m` `:. 24000= 600m` `:.m= (24000)/(600)= 40gm`