A) \[4.214{}^\circ C\]
B) \[42.14{}^\circ C\]
C) \[30{}^\circ C\]
D) None of these
Correct Answer: B
Solution :
[b] Energy supplied by the heater to the system in 10 min \[{{Q}_{1}}=P\times t=90J/s\times 10\times 60s\] \[=54000J=\frac{54000}{4.2}cal=12857cal\] Now if \[\theta \] is the final temperature of the system, energy absorbed by it to change its temperature from \[10{}^\circ C\]to \[\theta \] \[{}^\circ C\]is \[{{Q}_{2}}={{(ms\Delta T)}_{water}}+{{(ms\Delta T)}_{coil+calorimeter}}\] \[=360\times 1\times (\theta -10)+40(\theta -10)\] \[=400(\theta -10)\] According to problem, \[{{Q}_{1}}={{Q}_{2}}\] So \[12857=400(\theta -10)\]or \[\theta =42.14{}^\circ C\]You need to login to perform this action.
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