## How to calculate final temperature? (7.39)

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### How to calculate final temperature? (7.39)

A 50.0 g ice cube at 0.0 degree C is added to a glass containing 400 g of water at 45 degree C. What is the final temperature of the system? assume no heat is lost to the surroundings.

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### Re: How to calculate final temperature? (7.39)

This is how you set up the two sides of an equation. On one side, we need to melt the ice to liquid and then bring the liquid to final temperature. Since, melting ice and bringing the melted ice (liquid) to higher temperature require heat, it must come from somewhere. In this case, the amount of heat absorbed is from 400 g of liquid at 45 degrees C. This will cause the temperature of 400 g of liquid at 45 degrees C to go down. So, the other side of the equation represents the amount of heat released or transferred. Remember, since heat is released, the value is negative.

One side of the equation: Use $\Delta H_{fus}$ (on the back of Dr. Lavelle's periodic table) to calculate the amount of heat needed to melt 50.0 g of ice at 0 degrees C to 50.0 of liquid at 0 degrees C. Change grams to moles because $\Delta H_{fus}$ is per mole. Set this value aside for now. Use $q = nC\Delta T$. You have n and C. Heat capacity value can be found on the back of Dr. Lavelle's periodic table. In general, $\Delta T$ = ($T_{final} - T_{initial}$). So, your $\Delta T = (T_{final} - 0 degrees C)$. Finally, add these two values together.

On the other side of the equation. Use $q = n C\Delta T$. You have n and C. Your $\Delta T = (T_{final} - 45 degrees C)$. Remember to set this side to negative due to release of heat.

Now, solve for $T_{final}$.

Sandra
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### Re: How to calculate final temperature? (7.39)

When we are finding H(fusion). We change 50g of ice cube to moles. Do we also change 6.01kJ/mol to J/mol?

904286297
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### Re: How to calculate final temperature? (7.39)

Heat absorbed by ice cube = (50g/18.02g*mol )(6.01*10^3 J*mol^-1) + 50g*4.184J*C^-1 *g^-1 * (Final Temperature-0 degrees C)
Heat released by liquid water = (400g)(4.184J*C^-1*g^-1)(Final Temperature - 45 degrees C)
Heat absorbed + Heat released = 0
Solve for final temperature : T = (5.8 x 10^4 J)/(1.88 x 10^3 J*C^-1) = 31 degrees C

Twin 2
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### Re: How to calculate final temperature? (7.39)

So why would we not use the specific heat capacity of ice to calculate the heat of the ice cube? I'm confused because the answer uses specific heat capacity of liquid H20 which is 4.184 J/g C to find the heat of the ice cube.

y3chem
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### Re: How to calculate final temperature? (7.39)

The specific heat of water is used to calculate the heat of the ice cube melting because we are calculating the change in heat of the ice going from 0 to 45 degress after it melted into water. So we use the specific heat of liquid H2O to find the change of heat of the ice going from 0 to 45C.

y3chem
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### Re: How to calculate final temperature? (7.39)

Correction: Not to 45 degrees C. I meant to the final temperature

y3chem
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### Re: How to calculate final temperature? (7.39)

heat ice cube = (50g/18.02g*mol )(6.01*10^3 J*mol^-1) + 50g*4.184J*C^-1 *g^-1 * (Final Temperature-0 degrees C)

Essentially, if you take this equation apart
(50g/18.02g*mol )(6.01*10^3 J*mol^-1)
is calculating the heat absorbed by the ice cube as it melts into water (heat of fusion)

now ice is melted into water we use the specific heat of water (liquid H2O) to calculate the change in heat as the melted ice (=water) changes from zero degrees to its final temperature:
50g*4.184J*C^-1 *g^-1 * (Final Temperature)

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### Re: How to calculate final temperature? (7.39)

Sandra wrote:When we are finding H(fusion). We change 50g of ice cube to moles. Do we also change 6.01kJ/mol to J/mol?

The equation equates the energy absorbed by ice and energy released by liquid water. Since the unit of the specific heat capacity of liquid water is given as 4.184 J C-1 g-1, it will be more convenient to set the unit of energy on both sides of the equation the same. The conversion factor is 1kJ/mol=1000 J/mol

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### Re: How to calculate final temperature? (7.39)

Twin 2 wrote: So why would we not use the specific heat capacity of ice to calculate the heat of the ice cube? I'm confused because the answer uses specific heat capacity of liquid H20 which is 4.184 J/g C to find the heat of the ice cube.

The heat absorbed by the ice cube involves the heat absorbed by melting the ice at 0 oC to liquid water at 0 oC, and the heat absorbed by raising the same amount of liquid water at 0 oC to liquid water at the final temperature. Therefore, specific heat capacity of liquid water was used after the ice cube has melted.

DamianW
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### Re: How to calculate final temperature? (7.39)

When solving for T, do you ever multiply anything to Tfinal?