How Do You Calculate Heat Transfer with Limited Information?

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SUMMARY

The discussion centers on calculating heat transfer using the formula q = mcT, specifically addressing scenarios involving mass, temperature change, and heat exchange between substances. Participants clarify that the temperature change (ΔT) is the difference between final and initial temperatures, not the temperatures themselves. They emphasize the need to set up equations for both substances involved in heat exchange, ensuring that the heat lost by one equals the heat gained by the other. The discussion concludes with a definitive equation setup for calculating the final temperature when mixing gold and water.

PREREQUISITES
  • Understanding of the formula q = mcT for heat transfer calculations
  • Knowledge of specific heat capacities for different materials (e.g., gold and water)
  • Ability to set up and solve simultaneous equations
  • Familiarity with temperature change concepts in thermodynamics
NEXT STEPS
  • Research the specific heat capacity values for various materials
  • Learn how to derive and solve simultaneous equations in thermal equilibrium problems
  • Explore advanced heat transfer concepts, such as conduction and convection
  • Study real-world applications of heat transfer in engineering and environmental science
USEFUL FOR

Students in physics or chemistry, educators teaching thermodynamics, and professionals in engineering or environmental science who require a solid understanding of heat transfer principles.

Raerin
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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.
 
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Hi Raerin! :)

Raerin said:
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.
 
I like Serena said:
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)?
 
Raerin said:
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)
 

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