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Assume the enthalpy of water at 30° C to be 1446 kJ while its mass is 8 kg. The average specific heat capacity of is 4.2 kJ/kg.K. Calculate the enthalpy of water sample at 88 C.

Answer :

JannetPalos

Answer:

enthalpy of water at [tex]88^{0}\textrm{C}[/tex] is 3394.8 kJ

Explanation:

Mass does not change with change in temperature .

Hence change in enthalpy ([tex]\Delta H[/tex]) during increase in temperature from [tex]30^{0}\textrm{C}[/tex] to [tex]88^{0}\textrm{C}[/tex] is-

[tex]\Delta H=m\times c\times \Delta T[/tex]

where m is mass of water, c is specific heat capacity of water and [tex]\Delta T[/tex] is change in temperature.

Here, m = 8 kg, c = 4.2 kJ/(kg.K), [tex]\Delta T[/tex] = (88-30) K = 58 K

So, [tex]\Delta H=8kg\times 4.2 kJ/(kg.K)\times 58 K[/tex] = 1948.8 kJ

So, enthalpy of water at [tex]88^{0}\textrm{C}[/tex] = enthalpy of water at [tex]30^{0}\textrm{C}[/tex] + [tex]\Delta H[/tex] = (1446 + 1948.8) kJ = 3394.8 kJ

Answer:

enthalpy of water at  is 3394.8 kJ

Explanation:

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