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.…

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asked Dec 17, 2020 143k views

2 Answers

Junelle · Answer 1 · 2020-12-22T17:51:41+0000

Answer:

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