Find the temperature of the water after passing 5 g of steam at 100 °C through 500 g of water at 10 °C. Ans. (16.2°C) 

(Note: Specific heat of water is 4200 Jkg⁻¹K⁻¹, Latent heat of vaporization of water is 2.26 ｘ 10⁶ Jkg⁻¹).


Given:

Mass of the water  = m₁ = 500 g = 0.5 kg
Mass of the steam  = m₂ = 5 g = 0.005 kg
Temperature of water (initial Temperature) = Tₒ = T₁ = 10°C = (10 + 273) K = 283 K
Temperature of steam  = T₂ = 100°C = (100 + 273) K = 373 K
Specific heat of water  = c = 4200 J kg⁻¹K⁻¹
Latent heat of Vaporization of water = Hᵥ = 2.26 ｘ10⁶ Jkg⁻¹

To Find:

Final Temperature of Water  = T  = ?

Solution: 

Here the Final Temperature of Water  ( T ) can be found by using the formula of heat capacity.

Q  = m₁cΔT   (where ΔT = T - Tₒ)

Q  = m₁c (T - Tₒ)

or 

T  =  `\frac {Q}{m₁c}` + Tₒ    --------Eqn (1)


But the Heat Q is also unknown So, by using the formula for Latent heat of Vaporization

Q = m₂ Hv

By putting values 

Q = 0.005 kg ｘ 2.26 ｘ10⁶ J kg⁻¹

Q = 0.0113 ｘ10⁶ J

Q = 1.13 ｘ10⁴ J

Now using Eqn (1)

T  =  `\frac {Q}{m₂c}` + Tₒ  

by putting values 

T  =  `\frac {1.13 ｘ10⁴ J}{0.5 kg ｘ4200 J kg⁻¹K⁻¹}` + 283 K

T  =  0.000538 ｘ10⁴ K + 283 K

T  =  5.38 K + 283 K

T  =  288.38 K 

or

T  =  (288.38  - 273 )°C

T  =  15.38 °C

Thus the Final Temperature of the water is 15.38 °C

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