How Do You Calculate Heat Transfer with Limited Information?

In summary: And then you can solve for Tfinal. In summary, in order to solve these questions without knowing the initial or final temperatures, you can use the formula q = mcT and set up two equations for each substance involved. You then combine the equations and solve for the final temperature. The key is to consider the sign and magnitude of the heat q for each substance.
  • #1
Raerin
46
0
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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  • #2
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.
 
  • #3
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)?
 
  • #4
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)
 
  • #5
Without knowing the value of q, it is difficult to accurately calculate the final temperature. Additionally, the specific heat capacity of water changes with temperature, so a more precise calculation would require this information as well. It would be helpful to have more context or information about the problem in order to accurately solve it using thermodynamic calculations.
 

FAQ: How Do You Calculate Heat Transfer with Limited Information?

What is the purpose of thermodynamic calculations?

Thermodynamic calculations are used to determine the energy changes and equilibrium conditions of a system. They help scientists understand the behavior and properties of matter, and can be applied to a wide range of fields including chemistry, physics, and engineering.

What are some common thermodynamic variables used in calculations?

Some common thermodynamic variables include temperature, pressure, volume, and energy. These variables are often represented by the letters T, P, V, and E, respectively.

How are thermodynamic calculations used in real-world applications?

Thermodynamic calculations are used in a variety of real-world applications, such as designing efficient engines and power plants, predicting chemical reactions, and determining the energy efficiency of processes. They are also used in materials science to understand and improve the properties of materials.

What is the difference between thermodynamic calculations and thermodynamic laws?

Thermodynamic calculations involve the use of mathematical equations to determine the properties and behavior of a system. Thermodynamic laws, on the other hand, are fundamental principles that govern the behavior of energy and matter in a system. Thermodynamic calculations are based on these laws.

How can errors in thermodynamic calculations be minimized?

Errors in thermodynamic calculations can be minimized by using accurate and precise experimental data, using appropriate mathematical models and equations, and performing calculations multiple times to ensure consistency and accuracy. It is also important to understand the limitations and assumptions of the calculations being performed.

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