Need help on this question on Newtons law of cooling

In summary, the body will become more tempted after 12 minutes if it cools according to Newton's law of cooling.
  • #1
dan4loriel
8
0

Homework Statement



please i need help as have been trying to solve this question: A body cools from 70°C to 50°C in 6mins when the temp of the surrounding is 30°C. what wil be the tempt of the body after another 12mins if the cooling obeys Newtons law of cooling


equation:
dq/dt = -k(θ-θs)

where m- mass of the body
c-specific heat capacity of the body
θ-temp of the body
θs-temp of the surrounding or room

The Attempt at a Solution



i tried sovling it as follows:

dq=mc(70-50)

=20mc

dq/dt= 20mc/6= -k(70-50) =-20k

dq=mc(50-θs)
dq-dt=mc(50-θs)/12=-k

got lost from here please help!
 
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  • #2
Hi. by relation q=cθ the equation is

dθ/dt = -k/c (θ-θs) solving it
θ = (θ0-θs) e^-k/c t + θs
where θ0 is initial temperature of the body.

regards.
 
  • #3
thanks for the help
 
  • #4
hey,

wouldnt mind if you help me in solving them as regards to the equation given
 
Last edited by a moderator:
  • #5
dan4loriel said:

Homework Statement



please i need help as have been trying to solve this question: A body cools from 70°C to 50°C in 6mins when the temp of the surrounding is 30°C. what wil be the tempt of the body after another 12mins if the cooling obeys Newtons law of cooling


equation:
dq/dt = -k(θ-θs)

where m- mass of the body
c-specific heat capacity of the body
θ-temp of the body
θs-temp of the surrounding or room

The Attempt at a Solution



i tried sovling it as follows:

dq=mc(70-50)

=20mc

dq/dt= 20mc/6= -k(70-50) =-20k

dq=mc(50-θs)
dq-dt=mc(50-θs)/12=-k

got lost from here please help!

dan4loriel said:
hey,

wouldnt mind if you help me in solving them as regards to the equation given

We don't solve your homework problems for you here at the PF. We can give hints, like the one you got above. Now it is your turn to do wht work in solving the problem.

BTW -- I fixed up the "txt speak" abbreviation in your post. We do not allow txt speak abbreviations like "d" or "u" here. Please re-read the Rules link at the top of the page. Thanks.
 

1. What is Newton's law of cooling?

Newton's law of cooling is a physical law that states that the rate of heat loss of an object is proportional to the temperature difference between the object and its surroundings.

2. How is Newton's law of cooling used in real life?

Newton's law of cooling is used in various industries such as food and beverage, pharmaceutical, and HVAC systems to calculate the rate at which an object or substance cools down.

3. What is the formula for Newton's law of cooling?

The formula for Newton's law of cooling is dT/dt = -k(T - Tsurr), where dT/dt is the rate of change of temperature, k is the cooling constant, T is the temperature of the object, and Tsurr is the temperature of the surroundings.

4. How does Newton's law of cooling relate to the second law of thermodynamics?

Newton's law of cooling is a direct application of the second law of thermodynamics, which states that heat always flows from a higher temperature to a lower temperature. The law of cooling explains how this heat transfer occurs and can be used to predict the behavior of a cooling system.

5. Can Newton's law of cooling be applied to objects that are heating up?

Yes, Newton's law of cooling can also be applied to objects that are heating up. In this case, the rate of change of temperature would be positive, and the formula would be dT/dt = k(T - Tsurr), where k is the heating constant. This can be useful in predicting the behavior of objects that are being heated, such as in industrial processes or cooking.

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