Finding the Final Temperature of a Water Mixture: A Chemistry Problem

[wolfson_1123]

Hi all, i often help out in the chemistry section, was wondering if you could help me with a question:

A vessel of thermal capacity 500 J/oC contains 0.5 kg of water at 50oC.
0.2 kg of water at 20oC is added and perfectly mixed with the hotter water.
In this process, heat is lost by the hotter water and vessel to the colder water, so the hotter water and vessel cool down and the colder water heats up.
We will call the final temperature of the mixture (and vessel) ‘T’.

(a) Write an expression for the heat lost by the hotter water and vessel in terms of T.

(b) Write an expression for the heat gained by the colder water in terms of T.

(c) If we assume that no heat is lost from the vessel to the surroundings, the heat lost by the hotter water and vessel must be the same as the heat gained by the colder water. So by putting the above two expressions equal to each other, find the value of T.
The answer= 42.86oC


I know the answer if there it is just how to find it out. Thank you

 

[SGT]

The thermal capacity of water is 4.1868 J/g/⁰C, so your 500 g of hot water as a thermal capacity of 500*4.1868 J/⁰C.
Loss of heat by container plus hot water: (500*4.1868 + 500)*(50-T) J
Gain of heat by cold water: 200*4.18*(T-20)
Since both heat quantities are equal you can find T.

 

[wais]

for your question and explanation of the problem. Finding the final temperature of a water mixture involves understanding the principles of heat transfer and conservation of energy.

(a) To find the heat lost by the hotter water and vessel, we can use the formula Q = mcΔT, where Q is the heat lost, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature. In this case, the mass of the water is 0.5 kg and the change in temperature is (50-T)°C. The thermal capacity of the vessel is also given as 500 J/°C, so we can add this to the heat lost by the water. Therefore, the expression for the heat lost by the hotter water and vessel is Q = (0.5 kg)(4.18 J/g°C)(50-T)°C + (500 J/°C)(50-T)°C.

(b) Similarly, the heat gained by the colder water can be calculated using the same formula, but with a different mass and change in temperature. In this case, the mass is 0.2 kg and the change in temperature is (T-20)°C. Therefore, the expression for the heat gained by the colder water is Q = (0.2 kg)(4.18 J/g°C)(T-20)°C.

(c) To find the value of T, we can set these two expressions equal to each other and solve for T. This gives us:

(0.5 kg)(4.18 J/g°C)(50-T)°C + (500 J/°C)(50-T)°C = (0.2 kg)(4.18 J/g°C)(T-20)°C

Simplifying and solving for T, we get T = 42.86°C. This is the final temperature of the mixture and vessel.

I hope this helps you understand the problem better. Let me know if you have any further questions. Happy problem-solving!