Calculating the Entropy change of the universe [closed]

I often have trouble calculating the entropy change of the surrounding and almost always don't get the answer right for entropy of universe at the first attempt. I made an attempt for the following question to calculate the same but i am not sure if it is the right answer. I also don't have any answers to refer to. So Could someone tell me if what i have done is right and tell me the essential point to be kept in mind regarding the entropy of surroundings?

The question: An ice cube (m=5 gm) at a temperature of $-23C$ is dropped into a lake whose temperature is $27C$. After equilibrium is established, $\Delta S_{universe}$ is nearest to? $a)1.4\, b) 9.1\, c) 0.8\, d)8.2\, e)3.6$ all in $J/K$ [Latent heat of fusion $L_{ice} = 80 cal/g$, Specific heats $c_{water}=1\,c_{ice}=0.5$ (cal/gC)

1 Cal = 4.186 J

For system, (ice)

$\Delta S_{ice}= 5*0.5*4.186*lnT_{@0}/T_{@-23}+5*4.186*lnT_{@27}/T_{@0}+L*m*4.186/T_{@0}$

$= 9.02$

For surroundings,

$\Delta S_{surroundings}=-((mc*4.186*(T_@{0}-T_{@-23})+mc*4.186*(T_@{27}-T_{@0})+ L*m*4.186)/T_{@27}$

$=-8.26$

therefore $\Delta S_{universe} = 0.76$

nearest to option c.

closed as off-topic by ACuriousMind♦Apr 8 '17 at 11:51

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• The ice is definitely done correctly. But, for the lake water, I don't understand what the mc represents. If it is the initial mass of ice, then the first term should have a 0.5 on the heat capacity. – Chet Miller Apr 8 '17 at 11:44
• I'm voting to close this question as off-topic because check-my-work questions are off-topic. – ACuriousMind Apr 8 '17 at 11:51