A 0.598 g sample of steam at 104.8 ∘C is condensed into a container with 4.91 g of water at 16.7 ∘C. What is the final temperature of the water mixture if no heat is lost?

Question

and whoever answers this, can you please write where and how you got every number in the steps of the problem, please and thank you.

Answer 1

q = mC∆T

q = heat

m = mass

C = specific heat or heat capacity (C for steam is always 1.996 J/g°C; C for water is always 4.18 J/g°C)

∆T = change in temp

If no heat is lost, that means the heat lost by the steam is absorbed by the water.

Steam condenses at T = 100°C (it also evaporates at this temperature if it starts as a liquid).

So the steam is going to cool down to 100°C and then it becomes water. To calculate the total heat lost, we need to calculate the heat lost as it condenses to water (from 104.8° to 100° while it is still steam) and the heat lost as it cools in liquid form (from 100° to the final temperature).

So the total heat lost by the steam will be as follows:

q_steam = m_steamC_steam(100°C - T_steam) + m_steamC_water(T_final - 100°C)

m_steam = 0.598 g

C_steam = 1.996 J/g°C

T_steam = 104.8°C

C_water = 4.184 J/g°C

T_final = ? (what we are trying to solve for)

Because the steam is being cooled, the q will be a negative number (heat is lost). Let's plug in numbers:

q_steam = (0.598 g)(1.996 J/g°C)(100°C - 104.8°C) + (0.598 g)(4.184 J/g°C)(T_final - 100°C)

q_steam = -255.93 + 2.5T_final J

Now let's look at the heat gained by the water. The water is not changing phases, it is only changing temperature, so the heat gained is simply:

q_water = m_waterC_water(T_final - T_water)

m_water = 4.91 g

C_water = 4.184 J/g°C

T_final = ? (what we are trying to solve for)

T_water = 16.7°C

Plug in numbers:

q_water = (4.91 g)(4.184 J/g°C)(T_final - 16.7°C)

q_water = 20.54T_final - 343.075 J

We know the heat lost by steam will equal the heat gained by water. The q value for steam is negative, and the q value for water is positive, so:

-q_steam = q_water

-(-255.93 + 2.5T_final J) = 20.54T_final - 343.075 J

Now we can solve for T_final

T_final = 26°C

This makes sense because we know the water temperature will rise, and 26°C is greater than our starting temp of 16.7°C.

Something interesting to think about - the steam went all the way from over 100°C to about room temperature - that's a lot of heat lost! Why did the water only heat up by 10°C while the steam lost about 75°C?

Look at the heat capacities, it takes wayyy more energy (J) to heat or cool liquid water than steam (about 2 times as much energy per gram to heat or cool by 1°C), and look at the masses. We had way more water to heat up than steam to cool.

So steam doesn't take very much energy to heat up or cool down (specific heat) and we didn't have very much of it. 100% of the energy from condensing and cooling the steam went into a large amount of water that has a high heat capacity (requires a lot of energy to heat) which is why the temperature only went up a little.