Q. 54.0( 1 Vote )

# Indian style of c

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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