# Heat capacity / letent heat question

• blarginsnarf
Next can you tell me what the heat given off... by the cup or the water... is at the final temperature?The heat given off by the cup is 595 J. The heat given off by the water is 508 J.

## Homework Statement

An ice cube of mass 40g is transferred directly from the freezer where it is at -8 CELSIUS to a cup containing 120g of water at 95 degrees CELSIUS. Calculate the final temperature of water. Ignore any heat transfer to the cup or its surroundings

## Homework Equations

tf = ti + delta t

Is there a possibility elimination/substitution is involved? :S

## The Attempt at a Solution

0.04 (2.1 x 10^3)(delta t) + (0.12)(4.18 x 10^3)(delta t)

I solved for it and its wrong. The teacher gave us the answer, but not how to solve it so we can check it later. The answer is 50 degrees Celsius. Can someone show me how they solved it? :S

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Welcome to PF, blarginsnarf! blarginsnarf said:

## Homework Statement

An ice cube of mass 40g is transferred directly from the freezer where it is at -8 CELSIUS to a cup containing 120g of water at 95 degrees CELSIUS. Calculate the final temperature of water. Ignore any heat transfer to the cup or its surroundings

## Homework Equations

tf = ti + delta t

Is there a possibility elimination/substitution is involved? :S

## The Attempt at a Solution

0.04 (2.1 x 10^3)(delta t) + (0.12)(4.18 x 10^3)(delta t)

I solved for it and its wrong. The teacher gave us the answer, but not how to solve it so we can check it later. The answer is 50 degrees Celsius. Can someone show me how they solved it? :S

You are skipping a few steps here.
In particular you have not included the latent heat.

There is not just one "delta t" here, but there are 3 delta-t's.

First you have the delta t for the ice from its initial temperature to the melting point.
Second you have the delta t for the ice from the melting point to the as yet unknown final temperature.
And third you have the delta t for the hot water from its initial temperature to the unknown final temperature.

Can you write these 3 delta-t's in formula form?

I like Serena said:
Welcome to PF, blarginsnarf! You are skipping a few steps here.
In particular you have not included the latent heat.

There is not just one "delta t" here, but there are 3 delta-t's.

First you have the delta t for the ice from its initial temperature to the melting point.
Second you have the delta t for the ice from the melting point to the as yet unknown final temperature.
And third you have the delta t for the hot water from its initial temperature to the unknown final temperature.

Can you write these 3 delta-t's in formula form?

Alright, this is what I got:
1. delta t= 0 - (- 8)
delta t = 8
2. delta t = tf - 0
3. delta t = tf - 95

How's that? :S

Good! :)

What would be the heat absorbed by the ice (in formula form) to bring it first to the melting point, then melt it, and then bring it up to the final temperature?

I like Serena said:
Good! :)

What would be the heat absorbed by the ice (in formula form) to bring it first to the melting point, then melt it, and then bring it up to the final temperature?

Well according to my textbook, the heat capacity of ice is: 2.1 x 10 ^ 3

So... I did this: What would be the heat absorbed by the ice (in formula form) to bring it first to the melting point:

q = mct
q = 0.04 x 2.1 x 10 ^ 3 x 8
q = 672 J

then melt it:

Latent heat formula:
Latent heat of fusion for water in my textbook = 3.4 x 10 ^ 5

Qf = mlf
Qf = 0.04(3.4 x 10 ^ 5)
Qf = 13600 J

and then bring it up to the final temperature?:

I'm not sure at all :S

What I'm assuming: mass of ice + mass of water * latent heat of fusion for water? :S

blarginsnarf said:
Well according to my textbook, the heat capacity of ice is: 2.1 x 10 ^ 3

So... I did this: What would be the heat absorbed by the ice (in formula form) to bring it first to the melting point:

q = mct
q = 0.04 x 2.1 x 10 ^ 3 x 8
q = 672 J

Yep! blarginsnarf said:
then melt it:

Latent heat formula:
Latent heat of fusion for water in my textbook = 3.4 x 10 ^ 5

Qf = mlf
Qf = 0.04(3.4 x 10 ^ 5)
Qf = 13600 J

Good!

blarginsnarf said:
and then bring it up to the final temperature?:

I'm not sure at all :S

That would be, with a heat capacity of water of 4.2 x 10 ^ 3:
q = mcΔt
q = 0.04 x 4.2 x 10 ^ 3 x (tf - 0)

blarginsnarf said:
What I'm assuming: mass of ice + mass of water * latent heat of fusion for water? :S

Huh? I don't understand what you're saying here.
You're adding a mass to a heat? Next can you tell me what the heat given off by the hot water is (in formula form)?

The sum of the heats that the ice absorbs has to be equal to the heat the hot water gives off.
Can you set up the equation for that?

I like Serena said:
Yep! Good!

That would be, with a heat capacity of water of 4.2 x 10 ^ 3:
q = mcΔt
q = 0.04 x 4.2 x 10 ^ 3 x (tf - 0)

Huh? I don't understand what you're saying here.
You're adding a mass to a heat? Next can you tell me what the heat given off by the hot water is (in formula form)?

The sum of the heats that the ice absorbs has to be equal to the heat the hot water gives off.
Can you set up the equation for that?

Alright, so 672 + 13600 + 168tf
= 14272 + 168tf? :s I don't know tf

Next can you tell me what the heat given off by the hot water is (in formula form)?

I THINK it's like this:

Q = MCT
Q = 0.12 x 4.2 x 10 ^ 3 x(tf - 95)
Q = -46368 + 504tf
How's that? :D

blarginsnarf said:
Alright, so 672 + 13600 + 168tf
= 14272 + 168tf? :s I don't know tf

What's not to know?
That is right!

blarginsnarf said:
Next can you tell me what the heat given off by the hot water is (in formula form)?

I THINK it's like this:

Q = MCT
Q = 0.12 x 4.2 x 10 ^ 3 x(tf - 95)
Q = -46368 + 504tf
How's that? :D

Great!

Now the heat absorbed by the ice must be equal to the heat given off by the hot water.
Can you set them equal to each other (mind the minus sign) and solve for tf?

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Oh my gosh, I just solved it on paper. Thank you so much! You are a legend!

Thanx! 