Calculate the amount of heat required to convert 10 g ice at `-10^@C` into steam at `120^@C` at normal pressure. If heat is supplied uniformly at the rate `50cal//s`, sketch variation temperature with time

Explanation

Heat required to convert ice at `-10^@C` to ice at `0^@C`
`DeltaQ_1=ms_("ice")Deltatheta=10xx0.5[0-(-10)]=50cal`
Heat required to convert ice at `0^@C` to water at `0^@C`
`DeltaQ_2=mL_(ice)=10xx80=800cal`
Heat required to convert water at `0^@C` to water at `100^@C`
`DeltaQ_3=ms_omegaDeltatheta=10xx1xx[100-(0)]=1000cal`
Heat required to convert water at `100^@C` to steam at `100^@C`
`DeltaQ_4=mL_("steam")=10xx540=5400cal`
Heat required to convert steam at `100^@C` to steam at `120^@C`
`DeltaQ_5=ms_("steam")Deltatheta=10xx1xx[120-100]=200cal`
Total heat required
`DeltaQ=DeltaQ_1+DeltaQ_2+DeltaQ_3+DeltaQ_4+DeltaQ_5`
`=50+800+1000+5400+200`
`=7450cal`
Temperature v/s time graph:
Heat is supplied uniformly at the rate `50cal//s`
`DeltaQ=50t`
`-10^@C` to `0^@C: " "DeltaQ=ms_("ice")Deltatheta`
`50t=10xx0.5[theta-(-10)]`
`theta=-10+10t`
`theta` v/s t graph will be straight line with `+ve` slope and negative intercept.
When temperature remains same, `theta` v/s t graph will be horizontal line parallel to time axis.
When temperature increases. `theta` v/s t will be inclined straight line.