Heat Tranfer Problem with Varying Temperature

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The discussion revolves around a heat transfer problem involving a coffee cup exposed to cold temperatures. The initial energy loss rate from the cup is calculated at -4.424 J/s, using the formula for heat transfer. The challenge lies in estimating the coffee temperature after one minute of walking, considering the changing temperature of the coffee over time. Newton's Law of Cooling is suggested as a useful reference for understanding the heat transfer dynamics. The participant seeks assistance in applying this law to account for the varying temperature of the coffee during the walk.
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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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