How much heat energy is gained when 5, kg of | vedclass

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(A) The formula for heat energy gained is $Q = m \cdot c \cdot \Delta 	heta$.Given:Mass of water,$m = 5\, kg$.Specific heat of water,$c = 4.2\, kJ\,

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$1680$

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(A) The formula for heat energy gained is $Q = m \cdot c \cdot \Delta 	heta$.Given:Mass of water,$m = 5\, kg$.Specific heat of water,$c = 4.2\, kJ\, kg^{-1} {^{\circ}C}^{-1}$.Initial temperature,$	heta_1 = 20^{\circ}C$.Final temperature (boiling point),$	heta_2 = 100^{\circ}C$.Change in temperature,$\Delta 	heta = 	heta_2 - 	heta_1 = 100^{\circ}C - 20^{\circ}C = 80^{\circ}C$.Substituting the values into the formula:$Q = 5\, kg 	imes 4.2\, kJ\, kg^{-1} {^{\circ}C}^{-1} 	imes 80^{\circ}C$.$Q = 5 	imes 4.2 	imes 80\, kJ$.$Q = 21 	imes 80\, kJ$.$Q = 1680\, kJ$.

How much heat energy is gained when $5\, kg$ of water at $20^{\circ}C$ is brought to its boiling point? (Specific heat of water $= 4.2\, kJ\, kg^{-1} {^{\circ}C}^{-1}$)

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