Q 11-10-126JEE MainJEE Main 2017 (8 Apr)Easy
In an experiment, a sphere of aluminium of mass $0.20$ kg is heated up to $150^\circ$C. Immediately, it is put into water of volume $150$ cc at $27^\circ$C kept in a calorimeter of water equivalent to $0.025$ kg. The final temperature of the system is $40^\circ$C. The specific heat of aluminium is (take $4.2$ joule $= 1$ calorie):
Answer: (A) $434\ \text{J kg}^{-1}\,{}^\circ\text{C}^{-1}$
Water: $150$ cc $= 0.150$ kg; with the calorimeter's water equivalent the total is $0.175$ kg of water, warming by $13^\circ$C. The sphere cools by $110^\circ$C.
$$0.20\,c\,(110) = 0.175\times4200\times13$$
$$22c = 9555 \;\Rightarrow\; c \approx 434\ \text{J kg}^{-1}\,{}^\circ\text{C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics