$50 g$ 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 (in $^{\circ} C$ ) is:
(Specific heat of copper $= 420 J k g^{- 1} K^{- 1}$ ,
Specific heat of water $= 4186 J k g^{- 1} K^{- 1}$ )
5

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Solution

The correct option is A 5
Given:
Mass of copper, $m_{c} = 50 g = 0.05 k g$
Temperature change of copper coin, $Δ T_{c} = 10^{\circ} C$
Specific heat of copper, $S_{c} = 420 J k g^{- 1} K^{- 1}$
Mass of water, $m_{w} = 10 g = 0.01 k g$
Specific heat of water, $S_{w} = 4186 J k g^{- 1} K^{- 1}$ .

Let the temperature change of water be $Δ T_{w}$
Since, heat given to the copper is equal to the heat gained by the water,
$m_{c} S_{c} Δ T = m_{w} S_{w} Δ T_{w}$
$0.05 \times 420 \times 10 = 0.01 \times 4186 \times Δ T_{w}$
$Δ T_{w} = 5^{\circ} C$

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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, then the rise in its temperature (in ∘C ) is: (Specific heat of copper=420 J kg−1 K−1, Specific heat of water =4186 J kg−1 K−1)5

$50 g$ 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 (in $^{\circ} C$ ) is:
(Specific heat of copper $= 420 J k g^{- 1} K^{- 1}$ ,
Specific heat of water $= 4186 J k g^{- 1} K^{- 1}$ )
5