How much heat energy is gained when $0.6 k g$ of water at $25^{°} C$ is brought to its boiling point? [Sp. heat capacity of water $= 4200 J k g^{- 1} ° C^{- 1}$ ]

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Solution

Step 1: Given data
Mass $(m) = 0.6 k g$
Sp. heat capacity of water $(c) = 4200 J k g^{- 1} ° C^{- 1}$
Temperature $= 25 ° C$
Step 2: Calculating the heat absorbed by the water
We know, the boiling point of water is $100 ° C$ .
So, the rise in temperature $(θ_{R}) = (100 - 25) ° C = 75 ° C$
Now, heat gained $= m c θ_{R}$
$= 0.6 k g \times 4200 J k g^{- 1} ° C^{- 1} \times 75 ° C = 189000 J$
Therefore, the heat gained is $189000 J$ .

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The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 7,77,000 J. Calculate the specific latent heat of ice.

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Q. How much heat energy is gained when 5 kg of water at

20^{\circ} C

is brought to its boiling point?

Q. How much heat energy is gained when

5 k g

of water

20^{o} C

is brought to its boiling point (Specific heat of water

= 4.2 k J k g^{- 1}

Q. How much heat energy is gained when

5 k g

of water at

20^{\circ} C

is brought to its boiling point?

(Specific heat of water = 4.2 k J k g^{- 1} c^{- 1})

Q. How much heat energy is gained when

5 k g

of water at

20^{\circ} C

is brought to its boiling point?

(Specific heat of water = 4.2 k J k g^{- 1} c^{- 1})

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How much heat energy is gained when 0.6kgof water at 25°Cis brought to its boiling point? [Sp. heat capacity of water=4200Jkg-1°C-1]

How much heat energy is gained when $0.6 k g$ of water at $25^{°} C$ is brought to its boiling point? [Sp. heat capacity of water $= 4200 J k g^{- 1} ° C^{- 1}$ ]