(D) Given: Latent heat of vaporization $L_v = 22.6 \times 10^5 \ J \ kg^{-1}$.Mass of water,$m = 100 \ kg$.The heat required $(Q)$ to convert water in

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(D) Given: Latent heat of vaporization $L_v = 22.6 \times 10^5 \ J \ kg^{-1}$.Mass of water,$m = 100 \ kg$.The heat required $(Q)$ to convert water into vapour at constant temperature is given by the formula:$Q = m \times L_v$Substituting the given values:$Q = 100 \ kg \times (22.6 \times 10^5 \ J \ kg^{-1})$$Q = 10^2 \times 22.6 \times 10^5 \ J$$Q = 22.6 \times 10^7 \ J$Therefore,the amount of heat needed is $22.6 \times 10^7 \ J$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Latent Heat and Heating Curve

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Latent heat of vaporisation of water is $22.6 \times 10^5 \ J \ kg^{-1}$. The amount of heat needed to convert $100 \ kg$ of water into vapour at $100^{\circ} C$ is

TS EAMCET 2020 All TS EAMCET

A
$11.3 \times 10^5 \ J$
B
$11.3 \times 10^6 \ J$
C
$22.6 \times 10^6 \ J$
D
$22.6 \times 10^7 \ J$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Latent Heat and Heating Curve

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Latent heat of vaporisation of water is $22.6 \times 10^5 \ J \ kg^{-1}$. The amount of heat needed to convert $100 \ kg$ of water into vapour at $100^{\circ} C$ is

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