What is the final temperature when 8 kg of ice at -5 deg C is mixed with 2 kg of steam at 126.8 deg C.  The heat capacity of steam  is 1.901 kJ/kg-deg C and the heat capacity of ice is 2.108 kJ/kg-deg C.  The heat capacity of water is 4.183 kJ/kg-deg C. 

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This is a tricky problem because we do not know what the mixture of steam and ice look like after they come to the thermal equilibrium. Since there is such a large amount of ice (8 kg, compared to 2 kg of steam), let's calculate whether the heat given off...

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This is a tricky problem because we do not know what the mixture of steam and ice look like after they come to the thermal equilibrium. Since there is such a large amount of ice (8 kg, compared to 2 kg of steam), let's calculate whether the heat given off by the cooling and condensation of steam will be enough to melt all of the ice.

The heat that the steam gives off while cooling to 100 degrees C is

`Q_1 = c_sm_s(100 - T_(is)) = 1.901*10^3*2*26.8 = 1.019*10^5 J`

The heat that the steam gives off while condensing at the constant temperature of 100 degrees is

`Q_2 = L_v*m_s = 2.265*10^3*2 = 4530 J`

Here, `L_v` is the latent heat of vaporization of water/steam.` `

While the condensed steam cools off to 0 degree C, it will give off heat equal to

`Q_3 = c_w*m_s*(100 - 0) = 4183*2*100 = 8.366*10^5 J`

 

The total heat released by stem during these three processes is

`Q_1 +Q_2 +Q_3= 9.429*10^5 J` .

The heat it would take to melt 8 kg of ice at the temperature of 0 degrees C is

`Q_4= L_f*m_i = 334*10^3*8 = 2.672*10^6 J`

Here, `L_f` is the latent heat of fusion of water/ice. 

Since the heat released by steam is less than the heat required to melt all ice, only some of the ice will melt while the steam will condense and cool down to 0 degrees. Thus, the final temperature of the mixture will be 0 degrees C.

It can be calculated how much ice will melt, `Deltam` , from considering that the heat released by steam will heat up the ice to 0 degrees and melt some of the ice:

`Q_1 + Q_2 + Q_3 = c_i*m_i*(0 - T_(ii)) + L_f*Deltam`

`9.429*10^5 = 2108*8*5 +334*10^3*Deltam`

`Delta m = 2.57 kg`

The final temperature is 0 degree C, with 2.57 kg of the ice melted and all steam condensed to water.

 

 

 

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