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Specific Heat Capacity

50g of copper...

Question

$50 g$ of copper is heated to increase its temperature by $10^{0} C .$ If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is. $(s p e c i f i c h e a t o f c o p p e r = 420 J K^{- 1} ο C - 1)$

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 $5^{\circ} C$
Given,

Mass of copper $m_{c}$ = $50 g$

Rise in temperature $Δ T_{c}$ = $10^{\circ} C$

Mass of water $m_{ω}$ = $10 g$

Same amount of heat is supplied top copper and water; so

$m_{c} c_{c} Δ T_{c}$ = $m_{ω} c_{ω} Δ T_{ω}$

$\Rightarrow$ $(Δ T)_{ω}$ = $\frac{m_{c} c_{c} Δ T_{c}}{m_{ω} c_{ω}}$ = $\frac{50 \times 10^{- 3} \times 420 \times 10}{10 \times 10^{- 3} \times 4200}$ = $5^{\circ} C$

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Similar questions

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

50 g

of copper is heated to increase its temperature by

10^{0} C .

If the same quantity of heat is given to

10 g

of water, the rise in its temperature is.

(s p e c i f i c h e a t o f c o p p e r = 420 J K^{- 1} ο C - 1)

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, 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^{o}

C. If the same quantity of heat is given to

10

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

= 420

k g^{- 1} K^{- 1}

, specific heat of water

= 4200

k g^{- 1} K^{- 1}

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50g of copper is heated to increase its temperature by100C. If the same quantity of heat is given to 10g of water, the rise in its temperature is.(specific heat of copper =420 J K−1 οC−1)

The correct option is A 5∘CGiven, Mass of copper mc = 50 g Rise in temperature ΔTc = 10∘C Mass of water mω = 10 g Same amount of heat is supplied top copper and water; so mcccΔTc = mωcωΔTω ⇒ (ΔT)ω = mcccΔTcmωcω = 50 × 10−3 × 420 × 1010 × 10−3 × 4200 = 5∘C

$50 g$ of copper is heated to increase its temperature by $10^{0} C .$ If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is. $(s p e c i f i c h e a t o f c o p p e r = 420 J K^{- 1} ο C - 1)$