A) 7 kg
B) 6 kg
C) 4 kg
D) 2 kg
Correct Answer: B
Solution :
[b]: Heat released by 5 kg of water when its temperature fall from \[20{}^\circ C\] to \[0{}^\circ C\] is \[{{Q}_{1}}={{m}_{water}}{{s}_{water}}\Delta T\] \[=5\times {{10}^{3}}\times 1\times (20-0)={{10}^{5}}cal\] Heat absorbed by 2 kg of ice at \[-20{}^\circ C\] to increase its temperature to \[0{}^\circ C\]is \[{{Q}_{2}}={{m}_{ice}}{{s}_{ice}}\Delta T\] \[=2\times {{10}^{3}}0.5\times 20=0.2\times {{10}^{5}}cal\] So, the temperature of mixture will be \[0{}^\circ C\] \[0{}^\circ C\]. The remaining heat \[Q={{Q}_{1}}-{{Q}_{2}}=0.8\times {{10}^{5}}cal\] The remaining heat will melt a mass m of ice \[\therefore \]\[m=\frac{Q}{{{L}_{f}}}=\frac{0.8\times {{10}^{5}}}{80}={{10}^{3}}g=1kg\] The final mass of water in the container \[=\left( 5+1 \right)kg=6kg\]You need to login to perform this action.
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