Question

M grams of steam at 100°C is mixed with 200 g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40°C [heat of vaporization of water is 540 cal/ g and heat of fusion of ice is 80 cal/g], the value of M is______.

Answer 1

M_ice L_f + m_ice (40 – 0)C_W = m_steam L_v + m_steam (100 – 40)C_W

200[80 + 40(1)] = m[540 + 60(1)]

200(120) = m(600)

m = 40 gm

Answer 2

M × 540 + M + 60

= 200 × 80 + 200 × 1× (40– 0)

M = 40