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

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Infinity Learn · Accepted Answer

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 $$ice=0.5calorie/g/c$$  $$0$$  Δ$$T1=temperature$$ change from $$-10$$ to $$0=0-(-10)=10$$°$$CL1=latent$$ heat of fusion of $$ice=80$$ $$calorie/gC2=specific$$ heat of $$water=1calorie/g/c$$  $$0$$  Δ$$T2=temperature$$ change from $$0$$ to $$100=100-0=100$$°$$CL2=latent$$ heat of $$vaporisation=540$$ $$calorie/g$$  substitute above values in equation of Q  $$Q=mC1$$Δ$$T1+mL1+mC2$$Δ$$T2+mL2Q=1$$×$$0.5$$×$$10+1$$×$$80+1$$×$$1$$×$$100+1$$×$$540Q=5+80+100+540=725$$ cal.

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