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

Answer:

enthalpy of water at $88^{0}\textrm{C}$ is 3394.8 kJ

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-

$\Delta H=m\times c\times \Delta T$

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, $\Delta H=8kg\times 4.2 kJ/(kg.K)\times 58 K$ = 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

Answer 2

Answer:

enthalpy of water at  is 3394.8 kJ

Explanation: