Thermal physics help!

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#1
How would I do this MCQ?

You have to find the ratio between SHC and SLH?

But I’m not exactly sure how to pick up the value that is needed to plug into the formula?

Thanks
0
2 years ago
#2
(Original post by Yatayyat)
How would I do this MCQ?

You have to find the ratio between SHC and SLH?

But I’m not exactly sure how to pick up the value that is needed to plug into the formula?

Thanks
You have to assume the power stays constant so:
Q/T = mC^T/t and

^T/t=2 so Q/T = 2Mc = P

Also Q=Ml and P=Ml/t = 2Mc time for solidification is 20 minutes

l/20 = 2cso c/l=1/40
1
#3
(Original post by isiaiah d)
You have to assume the power stays constant so:
Q/T = mC^T/t and

^T/t=2 so Q/T = 2Mc = P

Also Q=Ml and P=Ml/t = 2Mc time for solidification is 20 minutes

l/20 = 2cso c/l=1/40
I’m confused to what formula you are even using? Is it Q = mc delta T and Q = ML?
0
2 years ago
#4
(Original post by Yatayyat)
How would I do this MCQ?

You have to find the ratio between SHC and SLH?

But I’m not exactly sure how to pick up the value that is needed to plug into the formula?

Thanks
Assume the rate of heat loss is constant and let rate of heat loss be P.

When the liquid is cooling, the rate of heat loss is

When the liquid is solidifying, the rate of heat loss is

Equate the two equations and you can find the required answer.
0
#5
(Original post by Eimmanuel)
Assume the rate of heat loss is constant and let rate of heat loss be P.

When the liquid is cooling, the rate of heat loss is

When the liquid is solidifying, the rate of heat loss is

Equate the two equations and you can find the required answer.
Shouldn't the second equation you put in be 'P = (M*L) / delta t, since power is the energy transferred per unit time
0
2 years ago
#6
(Original post by Yatayyat)
Shouldn't the second equation you put in be 'P = (M*L) / delta t, since power is the energy transferred per unit time
Yes, indeed. There are some typo in the tex.

P = \drac{m \times L \times}{\Delta t}

The correct tex should be

0
#7
Makes sense now. Cheers for all the help given.
0
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