Need help with Thermodynamics Q=mcΔt

[HollowScar]

Hi!

1. Homework Statement

A 3 kg block of ice at -15°C is placed into 30 kg of water at 323K. Calculate the final temperature.

It is simple, yet my answer doesn't match up.

2. Homework Equations

Oil flows through a heat exchanger at a flow rate of 125 L/min. It enters at 20°C and leaves at 75°C. The oil has a specific heat of 2.84 kJ/kgK and a relative density of 0.8. The oil is being heated by steam. The steam enters the heat exchanger at 100°C and the condensate formed leaves the heat exchanger at 85°C. Calculate the mass flow rate of the steam.

Any hints for the above question will definitely be appreciated.

3. The Attempt at a Solution

Specific heat of ice (snow) and steam = 2.04 kJ/kg°C
Specific heat of water = 4.183 kJ/kg°C
Latent heat of fusion = 335 kJ/kg
Latent heat of vaporization of water at 100°C = 2257 kJ/kg

Q(ice) = Q(water)
mcΔt = mcΔt
5 kg(2.04 kJ/kg°C)(85°C-(-10°C)) = m(2.04kJ/kg°C)(100°C-85°C)

Ans. = 1.53 kg, but does not match with mine.

Thank you.

 

[Stonebridge]

Q(ice) for the ice as it warms up, turns into water, and then warms up more, is the sum of 3 parts

The mass of ice, m, is 3kg (you have 5kg in your attempt)
part 1: mc_iceΔt for the ice warming from -15 to 0 deg C
part 2: mL for the ice melting
part 3: mc_waterΔt for the 3kg ice (now as water) warming up to its final temperature.

Q(water) is just mc_waterΔt where this is the 30kg of water cooling down to its final temperature.

 

[HollowScar]

Hi!

Thank you for the reply. Since ice, and steam eventually reach equilibrium, I would assume that I would have to use Latent heat of fusion, and vaporization. I will have to find the mass mass of steam involved.

 

[Stonebridge]

There won't be any steam in this question.
The initial temperature of the warm water is 323K, which is 50 deg C.
It cools down due to the ice being put in it.
In part 2 (in my answer) the value of L is the latent heat of fusion of ice.

 

[HollowScar]

There won't be any steam in this question.
The initial temperature of the warm water is 323K, which is 50 deg C.
It cools down due to the ice being put in it.
In part 2 (in my answer) the value of L is the latent heat of fusion of ice.

Sorry, I confused it with another question.

This is what I did.

mcΔt (ice) + mLf (ice) = mcΔt (water)

Temp of ice is (0-(-15C))
Temp of water is (49.85C-T)

I still did not get the right answer.

 

[Stonebridge]

You have not included part 3 of the 3 stages for the ice warming up.
When it has melted, part 2, it then warms up from zero C to some unknown temperature, call it T. (part 3)
The original warm water has to cool down to this temperature T from 50 deg C.

 

[HollowScar]

You have not included part 3 of the 3 stages for the ice warming up.
When it has melted, part 2, it then warms up from zero C to some unknown temperature, call it T. (part 3)
The original warm water has to cool down to this temperature T from 50 deg C.

mcΔt (ice) + mLf (ice) = mcΔt (water)

Temp of ice is (0-(-15C))
Temp of water is (T-50C)

Like this?

I know that the ice melts, and becomes more water. So only Latent Heat of Fusion is used.

 

[Stonebridge]

In part one that's correct
heat absorbed by ice = mc_icex15
[the temp rises 15 degs]

In part 2 the ice melts, that's also correct
heat absorbed by ice = mL [L=latent heat of fusion]
Part 3 is where the ice (now as water at 0 deg C) warms up to its final temperature T
heat absorbed by ice/water = mc_waterT
[the temp rises from 0 to T]

the total heat absorbed by the ice is the sum of these three [m=3kg]

For the initial mass of warm water [m=30kg]
the heat released = mc_waterx(temperature drop)
The water falls from 50 to T so the temperature drop is 50-T

Now combine all these into one equation and solve it for T

Heat absorbed by 3kg ice/water = heat released by 30kg water

 

[HollowScar]

Hi!

Sorry to bother you, and I really am indebted by your help, but I have problems understanding. Here is what I do.

Heat absorbed by 3kg ice/water = heat released by 30kg water
Thus mcΔt (ice) + mLf (ice) + mcΔt (water) = mcΔt (water)

However, the problem is somewhere above. I don't put the above in red, since I assume that the ice that melts automatically becomes water. It is easier that way since we don't know how much of the 3 kg block of ice melts into water. I just assume that ice melts, and it equals the 30 kg of water. If you could copy the above equation, and manipulate it, then it would be easy to understand. The answer of the final temperature is 37.5C.

I dislike thermodynamics. I have done fluid mechanics, and it was concrete in nature. This one takes assumptions. I am grateful for your help.

 

[Stonebridge]

Hi!

Sorry to bother you, and I really am indebted by your help, but I have problems understanding. Here is what I do.

Heat absorbed by 3kg ice/water = heat released by 30kg water
Thus mcΔt (ice) + mLf (ice) + mcΔt (water) = mcΔt (water)

That's fine.
You just need the values for the delta t.
1. ice warms up from -15 to 0 delta t=15
2. no change in temperature, ice turns to water, stays at 0 deg until all melted
3. water that was ice (m=3kg) warms up to T from zero, delta t = T-0 = T, the unknown final temperature.
4. the warm water cools down (it gives its heat to the ice then the 3kg of cold water) and falls to T. The final temperature of the whole mixture.
So delta t for the warm water is 50-T

Your equation becomes
3.c_ice.15 + 3.L_ice + 3.c_water.T = 30c_water.(50-T)

Can you find T now?

However, the problem is somewhere above. I don't put the above in red, since I assume that the ice that melts automatically becomes water. It is easier that way since we don't know how much of the 3 kg block of ice melts into water. I just assume that ice melts, and it equals the 30 kg of water. If you could copy the above equation, and manipulate it, then it would be easy to understand. The answer of the final temperature is 37.5C.

As the question asks for the final temperature, I assumed (correctly) that all the ice would melt, and then start to rise in temperature.
It could be possible, with more ice and less water at a lower temperature, that all the ice would not melt. That isn't the case here. If all the ice doesn't melt it's a trivial question because the final temperature would be zero. Water with ice in it.

I am grateful for your help.

You're welcome.

 

[HollowScar]

Thank you sir. It was really helpful. Got the answer right to the dot. One last question. So every time ice is mixed with water, or ice becomes water, do we have to also add the water which comes as a result of ice?

It is truly wonderful that I received help. I am glad, and indebted to your kindness.

 

[Stonebridge]

Thank you sir. It was really helpful. Got the answer right to the dot. One last question. So every time ice is mixed with water, or ice becomes water, do we have to also add the water which comes as a result of ice?

It is truly wonderful that I received help. I am glad, and indebted to your kindness.

Thank you.

These questions are always done the same way. It can seem strange at first.
The principle involved is one of conservation of (heat) energy.

We say that the system eventually reaches thermal equilibrium. Heat always flows from the hotter to the colder parts until this happens.
We then just equate the heat lost or rejected by the warmer parts, to the heat gained or absorbed by the colder parts.
In the case of simple mixing of two liquids at different temperatures, just use mcΔt for the one liquid = mcΔt for the other, making sure you identify the change in temperature, and remembering that the final temperature of both is the same at equilibrium.

You get m₁c₁ (T-T₁) = m₂c₂(T₂-T)

T₁ and T₂ are the initial temperatures of liquid 1 and 2 and T is the final equilibrium temperature. (Left side warmed up; right side cooled down.)
When you place ice in the water, you have to add to the left of that equation, the heat gained by the ice melting (m₁L) in order to balance the heat energy equation.
If the ice was at a lower temperature than 0, (T₃ say)then you also have to include on the left the energy given to it to rise first to 0 before it can melt, m₁c_iT₃ [I've ignored the minus sign for the temperature as the rise is 0 - (-T₃) which gives a plus.]
So the left of the equation is made up of 3 parts.

If there is not enough heat available in the warm water to melt the ice, then you finish up with some ice remaining and the mixture at 0 deg C.
What you would normally be asked to calculate is the amount of ice remaining, or the amount that melted. Just let mass of ice melted = m and equate the heat gained by that m kg of ice (mL) to that lost by the water in going from its original temperature to zero.