Steam is added to water at 49 degrees Celsius. What is the final temperature of the system?

Question

In a container of negligible mass, 0.045 kg of steam at 100 degrees C and atmospheric pressure is added to 0.170 kg of water at 49 degrees C. If no heat is lost to the surroundings, what is the final temperature of the system?

The specific heat of water is 4190 J/kg*K and the specific heat of steam is 2010 J/kg*K. The latent heat of vaporization for water is 2.26*10^6 J.

Answer 1

The specific heat of steam is irrelevant, as the steam is not changing temperature: it is merely condensing. This problem is much more tractable if worked in calories; the heat of vaporization of water is 540 calories per gram. If all the steam condenses, it releases 45 x 540 = 24300 calories. That much heat will raise the water temperature by 24300/170 = 142 C. Which is impossible, as it would raise the water temperature to over 100C. Hence, not all the steam condenses, and the final temperature will be 100 C. Note that this is true only if the excess steam is permitted to expand or escape; if that is not the case, one determines the amount of steam which must condense to raise the water temperature to 100 C, add that to the water initially present, and then do some fiddling with steam tables to get the final result.

Answer 2

We invoke the conservation of energy in that the heat energy before combining is equal to the heat energy after combining the steam and water.

Before combining, we have q = W w t; where W = .170 kg is the mass of water, w = 4190 J/kgK is the specific heat, and t = 49 degC the temperature; and Q = V v T; where V = .045 kg is the mass of the steam, v = 2010 J/kgK is the specific heat, and T = 100 degC is the temperature.

After combining, we have q' + C heat energy in the container of water; where C is the latent heat of condensation (vaporization). (Note I presumed you meant the latent heat and not the specific latent heat because of the units (J) you gave. If not, C = V c T ; where c is the specific latent heat would yield the latent heat.) C occurs and would be added to the resulting liquid water because the steam would condense as soon as its temperature dropped below 100 degC.

q' = the water heat for a mass W + V (we assume all the vapor/steam turned to liquid) = (W + V) w t'; and t' is the temperature of the water after combining.

Equating before and after, we have Q + q = q' + C; so that Q + q - C = q' = (W + V) w t' which gives us:

(Eqn 1) t' = [Q + q - C]/[(W + V) w];

and you have all the numbers to work this equation. You can do the math. Remember to convert degC to deg Kelvin (K = C + 273 deg) for unit consistency in Q = V v T and q = W w t

Physics lesson: Eqn 1 units are balanced, that is the RHS and the LHS have the same units of measure (deg K). So that's a good thing. Further, the equation is intuitively pleasing in that it says that for a given energy input (Q + q - C), the resulting temperature will be lower for more mass (W + V) of the resulting water. And that makes sense, because energy content depends on how much mass is at a given temperature. Finally, the latent heat of vaporization (condensation) had to be subtracted out of the input energy because it does not change temperature while the condensation is ongoing. [See source.]

Answer 3

Condensation of the steam = 0.045 x 2260
= 101.7kJ...added to 0.170kg water at 49°C = 49 + 17.3
= 66.3°C

0.045(100 - T) = 0.170(T – 66.3)
4.5 – 0.045T = 0.170T – 11.3
15.8 = 0.215T
Final temp. T = 73.5°C