1300 J of heat energy is supplied to raise the temperature of 0-5 kg of lead from 20- C to 40- C- Calculate the specific heat capacity of lead-

The correct option is C 130 J kg^ 1 K^ 1 As Heat Energy=ms(T_2 T_1) Specific heat is 13000.5×(40 20)=130010=130 Jkg^ 1K^ 1 ...

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Byju's Answer

Standard XII

Physics

Latent Heat

1300 J of hea...

Question

$1300 J$ of heat energy is supplied to raise the temperature of $0.5 k g$ of lead from $20^{\circ} C$ to $40^{\circ} C$ . Calculate the specific heat capacity of lead.

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

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

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

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1300 J k g^{- 1} K^{- 1}

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Solution

The correct option is C $130 J k g^{- 1} K^{- 1}$
As $H e a t$ $E n e r g y = m s (T_{2} - T_{1})$
Specific heat is $\frac{1300}{0.5 \times (40 - 20)} = \frac{1300}{10} = 130 J k g^{- 1} K^{- 1}$

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1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20∘C to 40∘C. Calculate the specific heat capacity of lead.

The correct option is C 130 Jkg−1K−1As Heat Energy=ms(T2−T1)Specific heat is 13000.5×(40−20)=130010=130 Jkg−1K−1

$1300 J$ of heat energy is supplied to raise the temperature of $0.5 k g$ of lead from $20^{\circ} C$ to $40^{\circ} C$ . Calculate the specific heat capacity of lead.

The correct option is C $130 J k g^{- 1} K^{- 1}$
As $H e a t$ $E n e r g y = m s (T_{2} - T_{1})$
Specific heat is $\frac{1300}{0.5 \times (40 - 20)} = \frac{1300}{10} = 130 J k g^{- 1} K^{- 1}$