The amount of heat required to convert 1 g of ice (specific 0.5 cal at $g^{- 1 o} C^{- 1}$ ) at $- 10^{0} C$ to steam at $100$ $^{\circ} C$ is ___________.
[ Given: Latent heat of ice is $80 C a l / g m,$ Latent heat of steam is $540 C a l / g m$ , Specific heat of water is $1 C a l / g m / C$ ]

Question

The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

]

Toppr Admin · Accepted Answer

Amount of heat required - Change the temp Of Ice from -10 C to 0 C - Heat required to melt the Ice - Heat required to increase the temperature of water from 0 to 100 C - Heat required to convert water-xA0- at 100 C to vapor at 100 C-1-xD7-0-5-0-x2212-x2212-10-1-xD7-80-1-xD7-1-xD7-100-1-xD7-540-5-80-100-540-725cal

The amount of heat required to convert 1 g of ice (specific 0.5 cal at g−1oC−1 ) at −100C to steam at 100 ∘C is ___________.[ Given: Latent heat of ice is 80Cal/gm, Latent heat of steam is 540Cal/gm, Specific heat of water is 1Cal/gm/C ]725 cal636 cal716 calNone of these

Amount of heat required = Change the temp Of Ice from -10 C to 0 C + Heat required to melt the Ice + Heat required to increase the temperature of water from 0 to 100 C + Heat required to convert water at 100 C to vapor at 100 C=1×0.5[0−(−10)]+1×80+1×1×100+1×540=5+80+100+540=725cal