# Heat and temperatureWatch

#1
A 500g ball of copper at 75celcius is cooled by dropping it into 300g of water at 20celcius. Assuming no heat is lost, what will be the equilibrium temperature? I don't have the answer to this question.. I have worked out to be 39.375celcius but im unsure if its right if somebody has different answer that they are definitely sure of can you post with a solution thanks..
0
5 years ago
#2
(Original post by Merdan)
A 500g ball of copper at 75celcius is cooled by dropping it into 300g of water at 20celcius. Assuming no heat is lost, what will be the equilibrium temperature? I don't have the answer to this question.. I have worked out to be 39.375celcius but im unsure if its right if somebody has different answer that they are definitely sure of can you post with a solution thanks..
We are not allowed to post complete solutions to numerical problems.
If you wish us to check your answer and method, post it here.
You also need to know the specific heat capacity of copper (and water) to do this.
#3
(Original post by Stonebridge)
We are not allowed to post complete solutions to numerical problems.
If you wish us to check your answer and method, post it here.
You also need to know the specific heat capacity of copper (and water) to do this.
Specific heat capacities are not stated in the question. Also if not allowed to post solutions I just claim the correct answer from someone who actually dealt with such a question before and say whether its right or wrong. If its wrong I will do extra work and search on internet for extra information since I know I did not answer it correctly... Plus people share all answers on maths forum of tsr go there and ban people ffs.
0
5 years ago
#4
(Original post by Merdan)
Specific heat capacities are not stated in the question. Also if not allowed to post solutions I just claim the correct answer from someone who actually dealt with such a question before and say whether its right or wrong. If its wrong I will do extra work and search on internet for extra information since I know I did not answer it correctly... Plus people share all answers on maths forum of tsr go there and ban people ffs.
The question cannot be done without the specific heat capacities of water and copper.
Using values off the internet I get 27 degC as the final temperature.
#5
(Original post by Stonebridge)
The question cannot be done without the specific heat capacities of water and copper.
Using values off the internet I get 27 degC as the final temperature.
Thanks sir when I used internet values I got 27.3celcius. The reason I got the temperature to be 39.375, I assumed that both specific heat capacities are the same so would cancel out and then did the calculation and I think I misunderstood the concepts specific heat capacity and Boltzmann constant. I somehow through they would be the same that's why I throught why they'd cancel out lel...
0
5 years ago
#6
(Original post by Stonebridge)
The question cannot be done without the specific heat capacities of water and copper.
Using values off the internet I get 27 degC as the final temperature.
Hey,

I understand heat involves the transfer of energy from a higher temperature to a lower one
but i don't exactly get what we are calling heat?
is it the quantity of energy transferred? so the amount of internal energy which has been removed by one object and hence gained by the other?

Thanks
0
5 years ago
#7
(Original post by flyhigh99)
Hey,

I understand heat involves the transfer of energy from a higher temperature to a lower one
but i don't exactly get what we are calling heat?
is it the quantity of energy transferred? so the amount of internal energy which has been removed by one object and hence gained by the other?

Thanks
Yes that's what it is in this question. The heat (internal energy) lost by the copper is equal to the heat gained by the water if none is lost to the surroundings. eg the container. It's just conservation of energy.
5 years ago
#8
(Original post by flyhigh99)
Hey,

I understand heat involves the transfer of energy from a higher temperature to a lower one
but i don't exactly get what we are calling heat?
is it the quantity of energy transferred? so the amount of internal energy which has been removed by one object and hence gained by the other?

Thanks
What people generally refer to as 'heat' is actually thermal internal energy. 'Heat' is not a property.

In other words, you have energy transfer by heating (formally correct term) from a higher temperature object to a lower temperature object (in your example). The delta T of each object, assuming no external heat losses, is the energy transferred by heating.

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