Certain amount of heat is given to 100 g of copper to increase its temperature by 21 °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-1 K-1 and that for water =4200Jkg-1ZK-1

Question

Certain amount of heat is given to 100 g of copper to increase its temperature by 21 °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-1 K-1 and that for water =4200Jkg-1ZK-1 (A) 4°C (B) 5.25°C (C) 8°C (D) 6°C

Tardigrade Admin · Accepted Answer

The amount of heat supplied is given by the
relation

Q = m c Δ T

Here , m = 100g = 0. 1 kg,c = 400J/ kg -K

Δ T = 21 K f or co pp er

T h u s, Q = 0.1 \times 400 \times 21 = 840 J

Hence ,

840 = 0.05 \times 4200 \times Δ T

\Rightarrow Δ T = 4^{\circ} C

Q. 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 $^{- 1} K^{- 1}$ and that for water =4200Jkg $^{- 1} Z K^{- 1}$

4 $^{\circ}$ C

5.25 $^{\circ}$ C

8 $^{\circ}$ C

6 $^{\circ}$ C

10.5 $^{\circ}$ C

The amount of heat supplied is given by the
relation $Q = m c Δ T$
Here , m = 100g = 0. 1 kg,c = 400J/ kg -K
$Δ T = 21 K f or co pp er$
$T h u s, Q = 0.1 \times 400 \times 21 = 840 J$
Hence , $840 = 0.05 \times 4200 \times Δ T$
$\Rightarrow Δ T = 4^{\circ} C$

Q. Certain amount of heat is given to 100 g of copper to increase its temperature by 21 ∘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 and that for water =4200Jkg−1ZK−1

4∘C

5.25∘C

8∘C

6∘C

10.5∘C

The amount of heat supplied is given by the relation Q=mcΔT Here , m = 100g = 0. 1 kg,c = 400J/ kg -K ΔT=21Kforcopper Thus,Q=0.1×400×21=840J Hence , 840=0.05×4200×ΔT ⇒ΔT=4∘C