Question

* (10) A thermally insulated pot has 150 g ice
at temperature (°C. How much steam of 100 °C
has to be mixed to it, so that water of
temperature 50°C will be obtained?
(Given : Latent heat of melting of ice = 80 cal/g,
latent heat of vaporization of water = 540 cal/g,
specific heat of water = 1 cal/g.°C)​

Answer 1

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  

Explanation: