# Help with A Level Physics questions (Specific heat capacity)

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Joe-G

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#1

Been trying to work out these questions for a while now and struggling Please can someone help me or point me in the right direction?

1) A flask contains 0.5kg of water at a temperature of 80 degrees Celsius. If 1.0kg of water at a temperature of 20 degrees Celsius is added to the flask what will be the temperature of the water in the flask after it has been mixed? You can assume that no heat is lost from the flask.

2) A copper poker with a mass of 1.0 kg and a temperature of 200 degrees Celcius is used to heat 200g of water. If the water has an initial temperature of 20 degrees Celcius what temperature might it be heated to by the poker? Explain why you are unlikely to obtain this temperature in practice.

Take specific heat capacity of:

water = 4200 J/Kg degrees Celcius

copper = 400 J/Kg degrees Celcius

1) A flask contains 0.5kg of water at a temperature of 80 degrees Celsius. If 1.0kg of water at a temperature of 20 degrees Celsius is added to the flask what will be the temperature of the water in the flask after it has been mixed? You can assume that no heat is lost from the flask.

2) A copper poker with a mass of 1.0 kg and a temperature of 200 degrees Celcius is used to heat 200g of water. If the water has an initial temperature of 20 degrees Celcius what temperature might it be heated to by the poker? Explain why you are unlikely to obtain this temperature in practice.

Take specific heat capacity of:

water = 4200 J/Kg degrees Celcius

copper = 400 J/Kg degrees Celcius

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Stonebridge

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#2

(Original post by

Been trying to work out these questions for a while now and struggling Please can someone help me or point me in the right direction?

1) A flask contains 0.5kg of water at a temperature of 80 degrees Celsius. If 1.0kg of water at a temperature of 20 degrees Celsius is added to the flask what will be the temperature of the water in the flask after it has been mixed? You can assume that no heat is lost from the flask.

2) A copper poker with a mass of 1.0 kg and a temperature of 200 degrees Celcius is used to heat 200g of water. If the water has an initial temperature of 20 degrees Celcius what temperature might it be heated to by the poker? Explain why you are unlikely to obtain this temperature in practice.

Take specific heat capacity of:

water = 4200 J/Kg degrees Celcius

copper = 400 J/Kg degrees Celcius

**Joe-G**)Been trying to work out these questions for a while now and struggling Please can someone help me or point me in the right direction?

1) A flask contains 0.5kg of water at a temperature of 80 degrees Celsius. If 1.0kg of water at a temperature of 20 degrees Celsius is added to the flask what will be the temperature of the water in the flask after it has been mixed? You can assume that no heat is lost from the flask.

2) A copper poker with a mass of 1.0 kg and a temperature of 200 degrees Celcius is used to heat 200g of water. If the water has an initial temperature of 20 degrees Celcius what temperature might it be heated to by the poker? Explain why you are unlikely to obtain this temperature in practice.

Take specific heat capacity of:

water = 4200 J/Kg degrees Celcius

copper = 400 J/Kg degrees Celcius

You should know the formula for heat lost or gained in terms of specific heat capacity, mass and temperature change.

What are you stuck on? What don't you understand?

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Joe-G

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#3

(Original post by

What are you stuck on? What don't you understand?

**Stonebridge**)What are you stuck on? What don't you understand?

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Stonebridge

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#4

(Original post by

I'm studying GCSE Physics atm but have been given these as an extra challenge. I know the formula but don't know how to apply it to this question

**Joe-G**)I'm studying GCSE Physics atm but have been given these as an extra challenge. I know the formula but don't know how to apply it to this question

Let the final temperature of the water be T (this is your unknown)

Write down the formula for the hot water as it cools to T from 80 degs

Write down the formula for the cool water as it heats up to T from 20 degs

Heat lost = heat gained if there is no loss.

Solve the equation for T

Same for the second question.

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Joe-G

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#5

(Original post by

Qu1

Let the final temperature of the water be T (this is your unknown)

Write down the formula for the hot water as it cools to T from 80 degs

Write down the formula for the cool water as it heats up to T from 20 degs

Heat lost = heat gained if there is no loss.

Solve the equation for T

Same for the second question.

**Stonebridge**)Qu1

Let the final temperature of the water be T (this is your unknown)

Write down the formula for the hot water as it cools to T from 80 degs

Write down the formula for the cool water as it heats up to T from 20 degs

Heat lost = heat gained if there is no loss.

Solve the equation for T

Same for the second question.

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Stonebridge

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#6

(Original post by

I don't get the heating and cooling part of using the equation. Could you quickly run through the first question please? I am using the equation E = m × c × θ. I have the mass and specific heat capacity but don't know what to use for the temperature.

**Joe-G**)I don't get the heating and cooling part of using the equation. Could you quickly run through the first question please? I am using the equation E = m × c × θ. I have the mass and specific heat capacity but don't know what to use for the temperature.

**change**in temperature in that formula.

If the hot water cools from 80 to theta, what is the change?

If the cool water heats from 20 to theta, what is the change?

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#7

(Original post by

theta is the

If the hot water cools from 80 to theta, what is the change?

If the cool water heats from 20 to theta, what is the change?

**Stonebridge**)theta is the

**change**in temperature in that formula.If the hot water cools from 80 to theta, what is the change?

If the cool water heats from 20 to theta, what is the change?

Let the final temperature of the water be T

Hot water as it cools to T from 80 degrees:

E = m x c x theta

= 0.5 x 4200 x 80

= 168 000

Cool water as it heats up to T from 20 degrees:

E = m x c x theta

= 1.5 x 4200 x 20

= 126 000

The formula for temperature would be Theta = E/mc but I'm lost before I've even got there, must have done something wrong

Thanks for such fast responses and all your help so far, nearly getting there

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#8

(Original post by

This is my working so far:

Let the final temperature of the water be T

Hot water as it cools to T from 80 degrees:

E = m x c x theta

= 0.5 x 4200 x 80

= 168 000

Cool water as it heats up to T from 20 degrees:

E = m x c x theta

= 1.5 x 4200 x 20

= 126 000

The formula for temperature would be Theta = E/mc but I'm lost before I've even got there, must have done something wrong

Thanks for such fast responses and all your help so far, nearly getting there

**Joe-G**)This is my working so far:

Let the final temperature of the water be T

Hot water as it cools to T from 80 degrees:

E = m x c x theta

= 0.5 x 4200 x 80

= 168 000

Cool water as it heats up to T from 20 degrees:

E = m x c x theta

= 1.5 x 4200 x 20

= 126 000

The formula for temperature would be Theta = E/mc but I'm lost before I've even got there, must have done something wrong

Thanks for such fast responses and all your help so far, nearly getting there

You have no

**T**in the formulas. And you need to calculate it!

If the water cools from 80 to T the change is not 80, is it? (That would be from 80 to zero.)

The change is 80 - T

If the temperature increases from 20 to T the change is not 20.

It would be T - 20

Now put these expressions for the change in temperature into the formula and solve for T

Can you do this now?

Equate heat lost to heat gained.

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#9

(Original post by

You have no

If the water cools from 80 to T the change is not 80, is it? (That would be from 80 to zero.)

The change is 80 - T

If the temperature increases from 20 to T the change is not 20.

It would be T - 20

Now put these expressions for the change in temperature into the formula and solve for T

Can you do this now?

Equate heat lost to heat gained.

**Stonebridge**)You have no

**T**in the formulas. And you need to calculate it!If the water cools from 80 to T the change is not 80, is it? (That would be from 80 to zero.)

The change is 80 - T

If the temperature increases from 20 to T the change is not 20.

It would be T - 20

Now put these expressions for the change in temperature into the formula and solve for T

Can you do this now?

Equate heat lost to heat gained.

= 2100 x (80-T)

E = 1.5 x 4200 x (T-20)

= 6300 x (T-20)

...

Ok I've given up :L This is in for tomorrow so I guess it doesn't matter now

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Jgfgjjfg

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#10

mate uwere nearly there so take that nd expND IT THEN SOLVE FOR Tcaps was accident sorry so the energy given from the hotter object is the same as the energy gained by the cooler object because it was assumed that no heat was lost therefore your two equationns are equal to each other as e is the same so you can equate them and find it. The reason that the hotter object is intial - final rather than the normal final - initial is because it is losing energy so e is a negative however that would not work mathematically to equate so we swap the numbers around to make it work mathematically if that makes sense?

i know i am very late-sorry about that

i know i am very late-sorry about that

Last edited by Jgfgjjfg; 2 years ago

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