$50 g$ 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°C

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6°C

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7°C

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8°C

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Solution

The correct option is A
5°C

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 = $420 J / k g /^{\circ} C$ , specific heat of water = $4200 J / k g /^{\circ} C$ )

50 g

of copper is heated to increase its temperature by

10^{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, then what would be the rise in its temperature?
(specific heat of copper $= 420 J / k g /^{\circ} C$ , specific heat of water $= 4200 J / k g /^{\circ} C$ )

$50 g$ 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, what will be the rise in its temperature? Assume no state change.
Specific heat of copper $= 420 J k g^{- 1 \circ} C$
Specific heat of water $= 4200 J k g^{- 1 \circ} C$

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