Temperature difference physics Problem

In summary, the temperature difference between a body and its surrounding medium decreases with time according to the equation T_B-T_M = (T_0-T_M)e^(-kt). From the given information, we know that after 5 minutes, the temperature difference decreases from 40°C to 20°C. The value of k is determined by solving 20= 40e^(-5k), and with that value, we can solve the equation 10= 40e^(-kt) to find the time it takes for the temperature difference to decrease to 10°C. The value of k is negative because the temperature difference is decreasing with time. Additionally, we can use the fact that the temperature difference halves every 5 minutes
  • #1
cscott
782
1
The temperature of a body differs from that of a medium, whose temperature is kept constant, by 40 °C. In 5 min, this difference is 20 °C. (a) What is the value of k? (b) In how many minutes will the difference in temperature be 10 °C?

[tex]T_B-T_M = \left ( T_0-T_M \right )e^{-kt}[/tex]

Do we know k is negative only because of what's said in part (b)?
 
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  • #2
I would say that the value of k is negative because it is the difference in temperature between the object and environment that is decreasing with time. It wouldn't make any difference whether the object was initially hotter or cooler than its surrrounding environment.
 
  • #3
Hammie said:
I would say that the value of k is negative because it is the difference in temperature between the object and environment that is decreasing with time. It wouldn't make any difference whether the object was initially hotter or cooler than its surrrounding environment.

Yeah, I should have quoted the part dealing with 5 minutes later. Thanks.
 
  • #4
cscott said:
The temperature of a body differs from that of a medium, whose temperature is kept constant, by 40 °C. In 5 min, this difference is 20 °C. (a) What is the value of k? (b) In how many minutes will the difference in temperature be 10 °C?

[tex]T_B-T_M = \left ( T_0-T_M \right )e^{-kt}[/tex]

Do we know k is negative only because of what's said in part (b)?

Actually we know that k is positive (so that -k is negative) because of the fact that the difference between the two temperatures is decreasing (from 40 to 20 in 5 minutes).

After 5 minutes, TB- TM= 20 and T0- TM= 40 so the equation, with t= 5 says
20= 40e-5k. It should be easy to solve that for k and then solve 10= 40e-kt, with that value of k, for t.

By the way, you don't really need to answer (a) in order to answer (b) (although you do have to answer (a) anyway!). If the temperature difference drops from 40 to 20 in 5 minutes, then it halves in in 5 minutes. In another 5 minutes, it will halve again! (That's a basic property of exponentials).
 

What is a temperature difference physics problem?

A temperature difference physics problem refers to a scenario where the temperature of two or more objects or systems is different, and the goal is to understand and analyze the changes in temperature or heat transfer between them.

What is the formula for calculating temperature difference?

The formula for calculating temperature difference is: ΔT = Tfinal - Tinitial, where ΔT is the temperature difference, Tfinal is the final temperature, and Tinitial is the initial temperature.

How is temperature difference related to heat transfer?

Temperature difference is directly related to heat transfer. Heat always flows from a higher temperature to a lower temperature, and the greater the temperature difference, the faster the heat transfer will occur.

What factors affect temperature difference in a physics problem?

The factors that affect temperature difference in a physics problem include the thermal conductivity of the materials involved, the surface area of the objects, the distance between the objects, and the duration of the heat transfer process.

How can temperature difference be measured and expressed?

Temperature difference can be measured using a thermometer or a thermocouple. It is typically expressed in units of degrees Celsius (°C) or Kelvin (K). In some cases, it may also be expressed in units of joules (J) or calories (cal) for heat transfer calculations.

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