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

[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.

 

[haruspex]

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?

 

[Chestermiller]

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

 

[haruspex]

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.

 

[Chestermiller]

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.

 

[JoeyBob]

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.

 

[haruspex]

-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.