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Physics Thermodynamic calculations

Raerin

Member
Oct 7, 2013
46
I know the formula q = mcT and that's the only formula I was taught before assigned these questions (in bold).

1. What mass of iron would be needed to have it increase by 6 degrees with 300J of heat.
I am not sure what is meant by the increase of 6 degrees. How do you solve it without knowing the final or initial temperatures?

2. If 10g of gold at 200 degrees Celsius is dropped into 100 grams of water at 20 degrees Celsius, what is the final temperature of the gold and the water?
How are you supposed to solve this without knowing the value of q?

3. If 100g of water at 25 degrees Celsius is mixed with 50g if water at 4 degrees Celsius, what is the final temperature?
I am confused about the same thing as in the second question.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,876
Hi Raerin! :)

I know the formula q = mcT and that's the only formula I was taught before assigned these questions (in bold).

1. What mass of iron would be needed to have it increase by 6 degrees with 300J of heat.
I am not sure what is meant by the increase of 6 degrees. How do you solve it without knowing the final or initial temperatures?
The T in your formula is the increase in temperature. It is not the initial nor the final termperature.
In other words, you do not need those temperatures.


2. If 10g of gold at 200 degrees Celsius is dropped into 100 grams of water at 20 degrees Celsius, what is the final temperature of the gold and the water?
How are you supposed to solve this without knowing the value of q?
You're supposed to set up 2 equations.
One for the heat q that is extracted from the gold, with the specific heat capacity c that belongs to gold
And one for the same heat q that is absorbed by the water, with the specific heat capacity c that belongs to water.


3. If 100g of water at 25 degrees Celsius is mixed with 50g if water at 4 degrees Celsius, what is the final temperature?
I am confused about the same thing as in the second question.
Works the same, except that both substances are water now.
 

Raerin

Member
Oct 7, 2013
46
Hi Raerin! :)
You're supposed to set up 2 equations.
One for the heat q that is extracted from the gold, with the specific heat capacity c that belongs to gold
And one for the same heat q that is absorbed by the water, with the specific heat capacity c that belongs to water.
That I understand but how am I supposed to find the final temperature without knowing q?

So for the gold:
q = 10g x 0.129 x (Tfinal - 200) degrees celsius

For water:
q = 100g x 4.184 x (Tfinal -20) degrees celsius

Am I supposed to supposed combine the equation as:
10g x 0.129 x (Tfinal - 200) = 100g x 4.184 x (Tfinal -20)?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,876
That I understand but how am I supposed to find the final temperature without knowing q?

So for the gold:
q = 10g x 0.129 x (Tfinal - 200) degrees celsius

For water:
q = 100g x 4.184 x (Tfinal -20) degrees celsius

Am I supposed to supposed combine the equation as:
10g x 0.129 x (Tfinal - 200) = 100g x 4.184 x (Tfinal -20)?
Almost!

The q for gold is negative, while the q for water is positive.
Their magnitude should be the same.
So:
10g x 0.129 x (200 - Tfinal) = 100g x 4.184 x (Tfinal - 20)