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Byju's Answer

Standard XII

Chemistry

Heat Capacity

50 g of coppe...

Question

50 g of copper is heated to increase its temperature by $10^{o} C$ . If the same quantity of heat is given to 10 g of water, the rise in temperature is $(s p e c i f i c h e a t o f c o p p e r = 420 J k g^{- 1} K^{- 1}, s p e c i f i c h e a t o f w a t e r = 4200 J k g^{- 1} K^{- 1})$

5^{o} C

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6^{o} C

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7^{o} C

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8^{o} C

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Solution

The correct option is A $5^{o} C$
Heat required
$= (50 \times 10^{- 3} k g) (420 J k g^{- 1} K^{- 1}) (10^{o} C)$
$= 210 J$
To produce 210 J by 10 g of water;temperature raised by t.
$∴ 210 = (10 \times 10^{- 3} k g) (4200 J k g^{- 1} K^{- 1}) (t)$
$t = \frac{210}{42} = 5^{o} C$

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Similar questions

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}

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}

)

$50 g$ 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 $= 4200 J k g^{- 1 \circ} C$

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

10^{o} C

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

=

420J/kg

^{o} C

; Specific heat of water

=

4200J/kg

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

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50 g of copper is heated to increase its temperature by 10oC. If the same quantity of heat is given to 10 g of water, the rise in temperature is (specificheatofcopper=420Jkg−1K−1,specificheatofwater=4200Jkg−1K−1)

The correct option is A 5oCHeat required=(50×10−3kg)(420Jkg−1K−1)(10oC)=210JTo produce 210 J by 10 g of water;temperature raised by t.∴210=(10×10−3kg)(4200Jkg−1K−1)(t)t=21042=5oC

50 g of copper is heated to increase its temperature by $10^{o} C$ . If the same quantity of heat is given to 10 g of water, the rise in temperature is $(s p e c i f i c h e a t o f c o p p e r = 420 J k g^{- 1} K^{- 1}, s p e c i f i c h e a t o f w a t e r = 4200 J k g^{- 1} K^{- 1})$

The correct option is A $5^{o} C$
Heat required
$= (50 \times 10^{- 3} k g) (420 J k g^{- 1} K^{- 1}) (10^{o} C)$
$= 210 J$
To produce 210 J by 10 g of water;temperature raised by t.
$∴ 210 = (10 \times 10^{- 3} k g) (4200 J k g^{- 1} K^{- 1}) (t)$
$t = \frac{210}{42} = 5^{o} C$