Heat Capacity & Temperature Changes

AI Thread Summary
The discussion focuses on understanding heat capacity and temperature changes in different materials. It highlights that adding the same amount of heat to different substances can result in varying temperature increases due to differences in mass and specific heat capacity. The correct formula for heat capacity is emphasized as Q = m × c × ΔT, and the calculations for a sample of aluminum indicate that 44,850 joules of heat is required to raise its temperature from 300K to 400K. Participants express uncertainty about their answers and seek clarification on the problems presented. Overall, the conversation aims to solidify understanding of heat transfer principles in physics.
MrScorpion
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Homework Statement



Quistions:

1-adding the same amount of heat to two samples does not produce the same increase in temperature. What physical parameters of hte bodies might be different?

2-We have a sample of metal with mass m. We supply an amount of heat Q and the temperature rises by deltaT.

(a) Write down the heat capacity of the sample in terms of m, Q and delta×T.

We increase the amount of heat supplied to the sample to 3Q

(b) What is the change in temperature?

(c)By what amount should we change the mass of the sample for the increase in temperature to be 10 deltaT?

3-We are going to increase the temperature of a sample of aluminium with mass 0.50kg. The initial temperature of the sampleis 300K. The final temperature is 400K. How much heat is required?

Homework Equations



1- (Hint: Q=m×c×deltaT)

3-(specific heat capacity of aluminium is 897 JKg^(-1)K^(-1)

The Attempt at a Solution



I don't know if the answers are correct or not

A1:The atoms. may be the first body is sold and the second bdy is liquid

A2:

(a) Q=m×deltaT

If we increase it to 3Q

==> 3Q=m×c×deltaT ==> Q=(m×deltaT)/3

(b) The change in temperature is deltaT = T2 - T1

(c) for this there is no attempt

A3:

Q=m×c×deltaT

==> 0.50 × 897 × (400 - 300 ) = 44850

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Please, please , please , and please i hope you to tell me what is the correct answer


i am trying allways to be a good student in physics but usually i don't answer quistions correctly.

please i hope you to teach me.

I will say thanks for you to the end of my life :smile:

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You have written a hint: Q = m*c*ΔT.
In the problem Q and ΔT remain constant. Then which quarantines should be different for the two bodies?
 
rl.bhat said:
You have written a hint: Q = m*c*ΔT.
In the problem Q and ΔT remain constant. Then which quarantines should be different for the two bodies?

First: thanks for your help

second:

The quarantines should be different for the two bodies are (c=heat capacity ,and m=mass of substance)

Is my answer right or not?



thanx again for your help:smile:
 
Your answer is correct.
 
rl.bhat said:
Your answer is correct.


Thank you

please:shy:, let's answer the remaining questions :redface: if you don't mind
 
MrScorpion said:
2-We have a sample of metal with mass m. We supply an amount of heat Q and the temperature rises by deltaT.

(a) Write down the heat capacity of the sample in terms of m, Q and delta×T.

We increase the amount of heat supplied to the sample to 3Q

(b) What is the change in temperature?

(c)By what amount should we change the mass of the sample for the increase in temperature to be 10 deltaT?

3-We are going to increase the temperature of a sample of aluminium with mass 0.50kg. The initial temperature of the sampleis 300K. The final temperature is 400K. How much heat is required?

Homework Equations



1- (Hint: Q=m×c×deltaT)

3-(specific heat capacity of aluminium is 897 JKg^(-1)K^(-1)

The Attempt at a Solution



I don't know if the answers are correct or not

A1:The atoms. may be the first body is sold and the second bdy is liquid

A2:

(a) Q=m×deltaT
No, Q=m×c×deltaT as stated in the hint. Therefore c=____?

If we increase it to 3Q

==> 3Q=m×c×deltaT ==> Q=(m×deltaT)/3
This goes with part (b), not (a).
(b) The change in temperature is deltaT = T2 - T1
You are not understanding what is being asked.
The temperature change is deltaT when the amount of heat supplied is Q. If, instead, an amount of heat 3Q is added, what is the temperature change then?
(c) for this there is no attempt

A3:

Q=m×c×deltaT

==> 0.50 × 897 × (400 - 300 ) = 44850
This looks correct.
 
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