$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is (specific heat of copper $= 420 J / k g /^{\circ} C$ )

5^{\circ} C

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6^{\circ} C

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7^{\circ} C

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8^{\circ} C

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Solution

The correct option is A $5^{\circ} C$
Heat absorbed by copper=heat absorbed by water
Heat absorbed= mass $\times$ specific heat $\times$ temperature difference
$\Rightarrow 50 \times 0.42 \times 10 = 10 \times 4.2 \times Δ T$
$\Rightarrow Δ T = 5^{o} C$

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$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is (specific heat of copper = $420 J / k g /^{\circ} C$ , specific heat of water = $4200 J / k g /^{\circ} C$ )

50 g

of copper is heated to increase its temperature by

10^{\circ} C

. If the same quantity of heat is given to

10 g

of water, then the rise in its temperature is (specific heat of copper

= 420 J / k g /^{\circ} C

)

$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, then what would be the rise in its temperature?
(specific heat of copper $= 420 J / k g /^{\circ} C$ , specific heat of water $= 4200 J / k g /^{\circ} C$ )

Q. 50 g of copper is heated to increase its temperature by

10^{\circ} C .

If the same quantity of heat is given to 10 g of water, the rise in its temperature is:
Specific heat of copper is

420 J k g^{- 1}^{\circ} C

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50g of copper is heated to increase its temperature by 10∘C. If the same quantity of heat is given to 10g of water, the rise in its temperature is (specific heat of copper =420J/kg/∘C)

The correct option is A 5∘CHeat absorbed by copper=heat absorbed by waterHeat absorbed= mass× specific heat× temperature difference⇒50×0.42×10=10×4.2×ΔT⇒ΔT=5oC

$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is (specific heat of copper $= 420 J / k g /^{\circ} C$ )

The correct option is A $5^{\circ} C$
Heat absorbed by copper=heat absorbed by water
Heat absorbed= mass $\times$ specific heat $\times$ temperature difference
$\Rightarrow 50 \times 0.42 \times 10 = 10 \times 4.2 \times Δ T$
$\Rightarrow Δ T = 5^{o} C$