Heat Tranfer Problem with Varying Temperature

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SUMMARY

The discussion focuses on calculating the rate of energy loss from a coffee cup in cold conditions and estimating the coffee temperature after one minute of walking. The coffee starts at 50°C, and the cup's thermal properties include a thickness of 2 mm and a thermal conductivity of 0.04 W m-1 K-1. The initial energy loss rate is calculated as -4.424 J/s using the equation Q = (-kAtΔT(t)/L. The challenge lies in accounting for the changing temperature of the coffee over time.

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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/Ldt + -kAt/LdT_{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.
 
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