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Physics Question on thermal properties of matter

Certain amount of heat is given to 100 g of copper to increase its temperature by 21 ∘^{\circ}C. if the same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat capacity of copper = 400 Jkg−1K−1^{-1} K^{-1} and that for water =4200Jkg−1ZK−1^{-1}ZK^{-1}

A

4∘^{\circ}C

B

5.25∘^{\circ}C

C

8∘^{\circ}C

D

6∘^{\circ}C

Answer

4∘^{\circ}C

Explanation

Solution

The amount of heat supplied is given by the
relation      Q=mcΔT \, \, \, \, \, Q=mc \Delta T
Here , m = 100g = 0. 1 kg,c = 400J/ kg -K
        ΔT=21K for copper\, \, \, \, \, \, \, \, \Delta T =21 K \, for \, copper
Thus,        Q=0.1×400×21=840JThus , \, \, \, \, \, \, \, \, Q = 0.1 \times 400 \times 21 = 840J
Hence ,      840=0.05×4200×ΔT \, \, \, \, \, 840=0.05 \times 4200 \times \Delta T
⇒          ΔT=4∘C\Rightarrow \, \, \, \, \, \, \, \, \, \, \Delta T =4^{\circ}C