Heat Tranfer Problem with Varying Temperature

[tejas710]

Homework Statement

1. Today it is your turn to go out and get coffee for your research group. It’s a cold January day (-20°C) and it takes 5 minutes to walk to the neighbourhood coffee shop. One of your group likes a large with double cream (75 ml). The coffee shop serves its coffee at 50°C and uses recyclable cups made from paper with thickness 2 mm and heat conductivity 0.04 W m-1 K-1. The large cup has a usable volume of 0.5 liter and a surface area of 316 cm2.
a) What is the rate of energy loss from the coffee cup as soon as you step out into the cold?
b) Estimate the temperature of the coffee after 1 minute of walking back to work. Assume
that coffee has the same heat capacity as water 4190 J kg-1 K-1

Homework Equations

Q = $$(-kAt\DeltaT (t)/L$$

The Attempt at a Solution

I have the solution to part a but I have no idea what to do for part b.
$$\partial Q$$/$$\partial t$$ = $$-kA(-20-T_{c})/L$$
$$\partial Q$$/$$\partial T_{c}$$ = $$-kAt/L$$
dQ = $$kA/L$$dt + $$-kAt/L$$$$dT_{c}$$
At t = 0, $$T_{c}$$ = 50
Therefore $$dQ/dt$$ = -4.424 $$J/s$$

But for part be I am not sure how to take the fact that $$T_{c}$$ keeps changing, thus the current and the total heat transferred as well. Any help on this would be greatly appreciated.

 

[LowlyPion]

I think you will have e^-kt relationship with the excess energy in your cup of coffee.

Here is a link to Newton's Law of Cooling:
http://en.wikipedia.org/wiki/Newton's_law_of_cooling#Newton.27s_law_of_cooling

It can explain things much better than I can.

 

[thomate1]

For part b, you can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the coffee, c is the specific heat capacity (in this case, it is the same as water, 4190 J kg-1 K-1), and ΔT is the change in temperature.

First, we need to calculate the mass of the coffee in the cup. The usable volume of the cup is 0.5 liters, which is equal to 500 ml. Since 1 ml of water has a mass of 1 gram, the mass of the coffee is 500 grams.

Next, we need to calculate the initial temperature of the coffee. Since the coffee shop serves its coffee at 50°C, the initial temperature is also 50°C.

Now, we can use the equation Q = mcΔT to calculate the heat transferred after 1 minute. Note that the change in temperature (ΔT) is equal to the difference between the final temperature and the initial temperature. So, after 1 minute, the final temperature is the temperature after 1 minute of walking in the cold. We can represent this as T_{f}.

Q = mc(T_{f}-50)

Now, we can use the equation for rate of energy loss from part a to calculate T_{f}.

\partial Q/\partial t = -kA(-20-T_{f})/L

\partial Q/\partial t = -4.424 J/s

k = 0.04 W m-1 K-1

A = 316 cm2 = 0.0316 m2

L = 0.002 m

Substituting these values into the equation and solving for T_{f}, we get T_{f} = 39.5°C.

Now, we can substitute this value into the equation for Q to calculate the heat transferred after 1 minute.

Q = mc(T_{f}-50)

Q = (0.5 kg)(4190 J kg-1 K-1)(39.5°C-50°C)

Q = -1957.5 J

Therefore, after 1 minute of walking, the coffee will have lost 1957.5 J of energy and its temperature will have decreased to 39.5°C.