## Trending News

# Advanced Physics (Thermodynamics) question?

I am having a difficult time with a certain exercise assigned to me. The question is as follows:

A house is at 20°C on a winter day when the outside temperature is a steady -15°C. The heat capacity of the house is 6.5 MJ/K and its thermal resistance is 6.67 mK/W. If the furnace suddenly fails (and there are no other sources of heat) how long will it take for the house temperature to reach the freezing point?

There is a hint with the question too:

Hint: Combine the differential forms of Equation 16.3 for heat capacity and Equation 16.5, in terms of thermal resistance, to show that the rate of temperature change is proportional to the temperature difference between the house and its surroundings. This relation is known as Newton's law of cooling.

Equation 16.3 for heat capacity is ΔQ=CΔT

Equation 16.5 in terms of thermal resistance is H=-ΔT/R

The answer to this problem is ~10 days.

My question is how do I get the differential forms of the above functions and how can I combine to create Newton's law of cooling.

This is what I think I have so far:

dQ=CdT

H=dQ/dt=-dT/R

When I combine these, the temperature cancels out which is not good, and when I integrate... well I am not sure what happens then. Any insight would be greatly appreciated!

Sorry the solution is ~10 hours

### 2 Answers

- EMLv 71 decade agoFavorite Answer
From dQ = CdT --> dQ/dt = C(dT/dt). [1]

The ΔT in H = -ΔT / R represents the temperature gradient between the house and the outside -- it's not a differential.

From H = dQ/dt = -ΔT / R --> -ΔT / R = C(dT/dt).

dT/dt = -ΔT / (RC)

dT/dt = -(T - Ta) / (RC), where Ta is the outside ambient temperature.

Solving the linear diff. equ.:

Integrating factor: e^(int[dt/(RC)]) = e^(t/(RC))

[T*e^(t/(RC))]' = e^(t/(RC))(Ta/(RC))

T*e^(t/(RC)) = int[e^(t/(RC))(Ta/(RC))*dt]

T*e^(t/(RC)) = Ta*e^(t/(RC)) + k

T = Ta + ke^(-t/(RC))

Solving for k using initial conditions, at t = 0, T = To:

To = Ta + k

k = To - Ta

T = Ta + (To - Ta)e^(-t/(RC))

Now we calculate the time needed for the house to reach 273K:

273 = 258 + (293 - 258)e^(-t/[(6.67e-3)(6.5e6)])

0.42857 = e^(-t/43355)

t = 36734.6 s --> 10.2 h

- alamilloLv 44 years ago
it somewhat is beneficial. do no longer journey undesirable. Ur smarter than those prevalent youthful ones. journey proud that ur gonna be working in a sturdy air conditioned place somewhat of mcdonalds merchandising burgers fries and apple pies.