Emily K.

asked • 08/02/16

Estimate it's temperature

A piece of metal is heated to 300 degrees Celsius and then placed in a cooling liquid at 50 degrees Celsius after 4 minutes, the metal has cooled 175 degrees Celsius estimate the temperature after 12 minutes
 

Steven W.

tutor
Don't feel bad, Emily.  Based on that information, I would be confused, too!  It almost looks like we are meant to assume a constant rate of temperature change versus time.  But, from a physics point of view, I can say that -- if the heat transfer from the solid metal is taking place mainly by conduction -- the heat transfer rate depends on the difference in temperature between the metal and the liquid.  So as the two approach each other in temperature, the rate at which heat is transferred (and thus the rate at which temperature changes at that point) slows down, and is definitely not linear.
 
Does this problem (or examples that are like it in your materials) given any indication of how they want you to express temperature change as a function of time in an equation?  Some indication of that (or a further discussion of the physics involved) would be necessary to figure out where to go with this problem.
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08/02/16

Kenneth S.

It looks to me as if you need to use Newton's Law of Cooling.  Consult your textbook or other class materials.
 
BTW...your title used it's but that should be its (possessive); reserve it's only for "it is"
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08/02/16

1 Expert Answer

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Arturo O. answered • 08/03/16

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Let us try to solve this using Newton's law of cooling, as Kenneth suggested:

T(t) = A + (T0 - A)*e^(-kt)

T0 = temperature of body at time t = 0, which is given as 300 C

t = time elapsed from moment when T = T0

k = constant with units of minutes^(-1)

A = temperature of the liquid bath, which we assume to be constant, given as 50 C.  (This assumes the liquid bath is a reservoir, i.e. massive enough so that its temperature remains the same during heat transfer).

First, we need to find k.

At t = 4 minutes, T(4) = 175 C
 
T(4) = A + (T0 - A)*e^(-kt) = 50 + (300 - 50)*e^(-4k) = 175
 
175 = 50 + (300 - 50)*e^(-4k) = 50 + 250*e^(-4k) 
 
125 = 250*e^(-4k) 
 
0.5 = e^(-4k) 
 
k = ln(0.5) / (-4) = 0.173287 minutes^(-1)
 
Then Newton's law of cooling for this problem becomes
 
T(t) = 50 + 250*e^(-0.173287 t)
 
with T in C, t in minutes.
 
T(12) = 50 + 250*e^[(-0.173287)(12)] ≅ 81.250 C
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