# Calorimetry - finding the final temperature of a system of ice and water

• JoeyBob
You should have J*K*L on the right.In summary, the water that cools down has a change in temperature of -21.3 kg*4.186 J*(x-346.95 K). This equation shows that the water has a mass of 7.1 kg*334000 J/kg and a temperature of 7.1 kg*4.186 J*(x-273.15 K).f

#### JoeyBob

Homework Statement
See attached
Relevant Equations
Q=mL

Q=mc(change T)
So all of the ice melts and I am guessing it then warms some so

Q=mL+mc(change in T)

for the water that cools down

Q=mc(change in T)

Q_cold = -Q_hot so -mc(Tf - Ti) = mL+mc(Tf - Ti)

My issue is that I have 2 unknowns. I don't know the specific heat capacity of water and I don't know the final temperature. I am not sure how I can find another equation to find the specific heat capacity of water either.

Even when I look up the specific heat capacity of water I am left with the wrong answer. specific heat capacity of water from internet is 4.186 J. When I use this I get -19346, which is obviously wrong. So I don't know if I am doing the equations wrong or if I need to find the specific heat capacity somehow.

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Homework Statement:: See attached
Relevant Equations:: Q=mL

Q=mc(change T)

specific heat capacity of water from internet is 4.186 J.
What units do you need it into match the other given numbers?

Please tell me you didn’t use the same value for Ti for water and ice.

Please tell me you didn’t use the same value for Ti for water and ice.
I'm pretty sure it's a units conversion issue.

• JoeyBob
What units do you need it into match the other given numbers?
What I was doing was converting Celsius to kelvin then converting it back into Celsius after I was done. Heres the link to the calculator I was using to check my algebra with the numbers plugged in. Probably easier to read but ill also type it out here.

https://www.symbolab.com/solver/algebra-calculator/-21.3\cdot4.186\left(x-346.95\right)=7.1\cdot334000+7.1\cdot4.186\cdot\left(x-273.15\right)

-21.3 kg * 4.186 J * (x-346.95 K) = 7.1 kg * 334000 J/kg + 7.1 kg * 4.186 J * (x-273.15)

The part on the left is of the relatively hot liquid water (mc(Tf-Ti)) where the part on the right is of the ice where it melts (mL+mc(Tf-Ti)).

As you can see I use their different masses and different initial temperatures for each.

What I was doing was converting Celsius to kelvin then converting it back into Celsius after I was done. Heres the link to the calculator I was using to check my algebra with the numbers plugged in. Probably easier to read but ill also type it out here.

https://www.symbolab.com/solver/algebra-calculator/-21.3\cdot4.186\left(x-346.95\right)=7.1\cdot334000+7.1\cdot4.186\cdot\left(x-273.15\right)

-21.3 kg * 4.186 J * (x-346.95 K) = 7.1 kg * 334000 J/kg + 7.1 kg * 4.186 J * (x-273.15)

The part on the left is of the relatively hot liquid water (mc(Tf-Ti)) where the part on the right is of the ice where it melts (mL+mc(Tf-Ti)).

As you can see I use their different masses and different initial temperatures for each.
That 4.18 is per gram, not per kilogram.

That 4.18 is per gram, not per kilogram.
Yeah I think that explains it. When I look up specific heat capacity of water 4.186 joules is in bold. You have to read below to see 4.186 J/g. Idk why the correct units arent displayed initially.

-21.3 kg * 4.186 J * (x-346.95 K) = 7.1 kg * 334000 J/kg + 7.1 kg * 4.186 J * (x-273.15)
Take a good look at all those units.
On the left you have kg*J*K.
First term on the right is kg*J/kg = J (so that one is right).
Second term on the right you show as kg*J.

• JoeyBob