Answer to Question #155907 in Mechanics | Relativity for Emmanuel

Question

Dry steam at 100°C is bubbled into 500g of water originally at 20°C. What will be the temperature of the water after 30g of steam has condensed? Specific latent heat of steam is 2.26*10^6J/k, specific heat capacity of water is 4200J/Kg/k

Answer 1

As the system is supposed to be isolated, the total heat flow is zero:

$Q_{steam}+Q_{water}=0$

The heat transferred by steam is related to it's condensation, so we will count it as negative. The heat recieved by water is related to it's heating, so we will count it as positive. Thus we have :

$-Lm_{cond.steam} + cm_{water}(T_{final}-T_{initial}) = 0$

$T_{final} =\frac{Lm_{steam}}{cm_{water}} +T_{initial}$

$T_{final}=\frac{2.26\cdot10^6\cdot 3\cdot10^{-2}}{4.2\cdot10^3\cdot 0.5}+20 = 52.3 ^{\circ}C$