M grams of steam at 100^ C is mixed with 200g 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…

Question

M grams of steam at 100^ C is mixed with 200g 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

Accepted Answer

Correct answer: 40. (40) M_ iceL_f+m_ ice (40-0 )C_w=m_ steamL_v+m_ steam (100-40 )C_w ⇒200 [80+40 (1 ) ]=m [540+60 (1 ) ] ⇒200 (120 )=m (600 ) m=40gm

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

Solution

More Heat Transfer PYQs