Question

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 1

enthalpy of water at is 3394.8 kJ

Step-by-step explanation:

Answer 2

enthalpy of water at
`88^(0)\textrm{C}` is 3394.8 kJ

Step-by-step explanation:

Mass does not change with change in temperature .

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

`\Delta H=m* c* \Delta T`

where m is mass of water, c is specific heat capacity of water and
`\Delta T` is change in temperature.

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

So,
`\Delta H=8kg* 4.2 kJ/(kg.K)* 58 K` = 1948.8 kJ

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