Entropy and Carnot refrigerators

1. The problem statement, all variables and given/known data
a)A stone at 400K with heat capacity ##c_p## is placed in a very large lake at 300K. The stone cools rapidly to 300K. Calculate the entropy change, due to this process, of the stone and lake.

b)An insulated cool-box of a Carnot refrigerator at temperature T loses heat at a rate proportional to the difference between T and the atmosphere To, i.e ##\dot{Q} = B(T_o-T)## where B is some constant. Show that the temperature of the cool-box is given by the solution of ##T^2 – \left(2T_o + \frac{W}{B}\right)T + T_o^2 = 0##

2. Relevant equations
Definition of entropy ##dS = dQ/T##. Carnot refrigerator efficiency

3. The attempt at a solution
a)The stone may be regarded as the system and the lake as the surroundings. The lake is effectively a reservoir and so the addition of the stone to the lake will alter the temperature of the lake negligibly. When the stone is placed in the lake, there will be a net flow from lake → stone until they come to thermal equilibrium whereby the heat flow is equilibrated.
##dS_{stone} = \frac{dQ_{res}}{T} = \frac{c_p dT}{T}##. ##dQ_{res}## is the amount of heat the reservoir (or lake) provides to the stone to change its temperature to 300K. Evaluating gives ##\Delta S = c_p \ln(3/4) < 0## as expected. Intuitively, the entropy of the reservoir should change negligibly, but I don’t think I am supposed to ignore such a change.

By the second law, ##\Delta S_{stone} + \Delta S_{lake} \geq 0## and in this case, no equality since the process is irreversible. So ##\Delta S_{lake}## must be positive to accommodate this. (albeit very small – the addition of the rock increases the entropy of the lake not significantly). How would I calculate this?

b)My starting point was to use the Carnot efficiency ##Q_1/(Q_1 – Q_2) = T_1/(T_1-T_2)##, where ##Q_1## is the heat taken from the refrigerator and ##Q_1-Q_2## is the work provided to run the refrigerator. But I am not sure how to use this and/or the expression given.
Thanks.

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