Question #1a163

1 Answer
A08
Apr 25, 2016

#154500cal#

Explanation:

We know that heat gained/lost is given by
#DeltaQ=mst#, or #DeltaQ=mL#
where #m,s and t# are the mass, specific heat and rise or gain in temperature of the object;
#L# is the latent heat for the change of state.

In the given problem heat is given to water to increase its temperature from #22^@"C"# to #100^@"C"# and thereafter boil the water to make it steam at #100^@"C"#.

  1. Water heated from #22^@"C"# to #100^@"C"#
    Using the expression and taking specific heat of water as 1.
    #DeltaQ=mst#
    #DeltaQ_1=250xx1xx(100-22)=19500cal#

  2. Heat gained by boiling water to change into vapours at the same temperature i.e., #100^@"C"# is given by #DeltaQ_2=mL#, where Latent heat of vaporization of water is #540calg^-1#.
    #:. DeltaQ_2=250xx540=135000cal#

Total heat required #=DeltaQ_1+DeltaQ_2=19500+135000#
#=154500cal#

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