A) \[0{}^\circ C\]
B) \[40{}^\circ C\]
C) \[80{}^\circ C\]
D) less than \[0{}^\circ C\]
Correct Answer: A
Solution :
Heat taken by ice to melt at \[{{0}^{o}}C\] is \[{{Q}_{1}}=mL=540\times 80=43200\,\,cal\] Heat given by water to cool upto \[{{0}^{o}}C\] is \[{{Q}_{2}}=m\times s\times \Delta \theta \] \[=540\times 1\times (80-0)\] \[=43200\,\,cal\] Heat given by water = Heat absorbed by ice \[\therefore \]Three will be no change in the temperature of mixture. Hence, the temperature of mixture will be\[{{0}^{o}}C\].
You need to login to perform this action.
You will be redirected in
3 sec
