Answer :

Given


Specific heat capacity of water = 4200 J kg–1 °C–1


latent heat of vaporization of water = 2.27 × 106 J kg–1.


Formula used:


ΔQ=MCΔT


ΔQ=heat exchange


ΔT=tempeature change


M=mass


C=heat capacity


Heat lost per second from the atmosphere = Heat gained by the water


0.2× 10-3× 2.27× 106=4200×(10-0.2× 10-3)× ΔT


ΔT=0.01080974° C


Therefore time required for 5° change=5/0.01080974=462.5 seconds=7.7 minutes


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