Questions

# The amount of heat required to convert 1 gm of ice at -10° C to steam at 100° C is :

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By Expert Faculty of Sri Chaitanya
a
725 cal
b
1000 cal
c
800 cal
d
80 cal
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detailed solution

Correct option is A

heat required Q=Q1+Q2+Q3+Q4total amount of heat required is Q=Q1+Q2+Q3+Q4here Q1= amount of heat required to convert ice at -100C to ice at 00C=mC1∆T1Q2= amount of heat required to convert ice at 00C to water at  00C =mL1 Q3= amount of heat required to convert water at  00C to water at  1000C=mC2∆T2Q4= amount of heat required to convert water at  1000C to steam at  1000C=mL2here m is mass =1gmC1=specific heat of ice=0.5calorie/g/c  0  ΔT1=temperature change from -10 to 0=0-(-10)=10°CL1=latent heat of fusion of ice=80 calorie/gC2=specific heat of water=1calorie/g/c  0  ΔT2=temperature change from 0 to 100=100-0=100°CL2=latent heat of vaporisation=540 calorie/g  substitute above values in equation of Q  Q=mC1ΔT1+mL1+mC2ΔT2+mL2Q=1×0.5×10+1×80+1×1×100+1×540Q=5+80+100+540=725 cal.

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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