How much hot water is needed to bring the bath back to its original temperature?

Question

A bath is filled up with hot water, say 42 degrees C. A volume of cold water is added, taking the temperature down to 38 degrees C.

As a proportion of the water that is already in the bath, how much hot water (42 degrees C) will be needed to bring the whole bath back up to 42 degrees C.

If you could show your working and explain any laws you are using that would be great.

Answer 1

Use Q=mcθ
where Q=energy in Joules
m=mass in kg
c=specific heat capacity, unit is J/[(kg)(℃)]
specific heat capacity of water is 4200 (around that)
temperature change is 4 deg
assume mass of water at first is 100g.
assuming no heat loss, the loss of heat in the water of 42deg = gain of heat in the water of 38 deg.
thus Q1 = Q2.
(100)(4200)(4)=(4200)(4)m
m=0.01g
thus the total mass of water now is 100.001g.
now, to pour 42deg water to bring it up to 42deg.
Q1=Q2
m1c1θ1=m2c2θ2
(100.001)(4200)(4)=m(4200)(0)-->no change in temp. stay at 42
thus, m=[(100.001)(4200)(4)]/[(4200)(0)]
the answer is undefined , ∞
therefore, the amount of water is undefined, or rather, infinity. impossible to bring it back up to 42 deg with water of temp 42 deg.

remember, eqn is Q=mcθ
where Q=energy in Joules
m=mass in kg
c=specific heat capacity, unit is J/[(kg)(℃)]

Answer 2

Considering the system of is an ideal system. There is no heat loss through the walls of the container - the tub. And there is no heat loss to the atmosphere from the water while it is being added to the tub.

The second consideration is a question about the size of the tub. If the tub is limitless in volume then a limitless amount of hot water can be added after cold water has been added. In that case more hot water at 42 degrees is added the less will be the cooling affect of the cold water previously added and so on. Ideally it is impossible to reach the same temperature – 42 degrees – does not matter how much more hot water is added. Only a theoretical answer can be found through applying Calculus methods – Integration method. I am unfortunately our practice for some time now therefore I can only suggest a scenario and cannot show you the workings of and calculation. But I think this can be done.

If on the other hand the container is the tub of an ordinary measurable size and water is not to be spilled that it is impossible to cancel the cooling effect of the cold water added.

Answer 3

You would never bring the temperature back up to 42 degress however much water you added at that same temperature. As you added more and more, then even if the water was not already losing heat all the time anyway (which it is, but we will ignore that for the sake of argument), all that will happen is that the temperature will get closer and closer to 42, never quite reaching it.

Answer 4

infinite hot water. this is because the water you're adding is not replacing the colder water in the bath there for the water's temp would get increasingly closer to 42 degrees but never reach it also the water would be cooling as you poured it so this would not be very good. the only way of doing it would be to use water that was hotter than the original temp. to cancel out the heat loss total. hope this was helpfull

Answer 5

An infinite amount.

The overall temperature of the water will always be somewhere between the starting temperature of the bath and the temperature of the water added to it. You need to add water hotter than the desired temperature.

Answer 6

if no heat lost during the process, add 1/10 of total volume of hot water.

the temp. may be 50.


or equal hot water(100) to amount of cold water(0) added before

Answer 7

what is the temperature of the cold water added?

Answer 8

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Answer 9

Boyle's Law
P1.V1 over T1 = P2.V2. over T2
Transpose and finish.

You work it out and you might remember how to use it in future.

Answer 10

Wow. Iv seen some crazy questions but this one is bizarre