Maharashtra Class 10 Answered
A themally insulated pot has 150g ice at temperature 0 c. how much steam of 100 degree celcius has to be mixed to it so that water of temperature 50 degree celcius will be obtained?
Asked by kartikkhedkar4 | 11 Jul, 2019, 08:23: PM
Expert Answer
m1 = 150 g,
ΔT1 = 50 - 0 = 50 °C
specific heat capacity of water, c = 1 cal/g °C
Latent heat of vapourisation of water = 540 cal/g
Latent heat of melting of ice = 80 cal/g
Heat absorbed by ice, Q1 = m1L1 = 150 x 80 = 12000 cal
Heat absorbed by water formed by meting of ice, Q2 = m1 c ΔT1 = 150 x 1 x 50 = 7500 cal
Heat released by the steam, Q3 = m2L2 = m2 x 540
Heat released by water formed by condensation of steam, Q4 = m2 c T2 = m2 x 1 x 50
According to principle of heat exchange,
Heat absorbed = Heat released
Q1 + Q2 = Q3 + Q4
12000 + 7500 = m2 × 540 + m2 x 50
19500 = m2 (540+50)
m2 = 33.05 g
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