The change of energy on vaporizing 1.00 kg of liquid water at 0° C and 1 atm is

Question

The change of energy on vaporizing 1.00 kg of liquid water at 0° C and 1 atm is (A)  2367 kJ kg-1  (B)  -2367 kJ kg-1 (C)  -2367 kJ mol-1  (D)  -2367 kJ g-1

Tardigrade Admin · Accepted Answer

The change of energy on vaporizing 1.00 kg of any liquid at 0°C and 1 atm is called its latent heat of vaporization. For water its value is 2367 kJ kg-1.

Q. The change of energy on vaporizing $1.00 k g$ of liquid water at $0^{\circ} C$ and $1 a t m$ is

$2367 k J k g^{- 1}$

$- 2367 k J k g^{- 1}$

$- 2367 k J m o l^{- 1}$

$- 2367 k J g^{- 1}$

The change of energy on vaporizing $1.00 k g$ of any liquid at $0^{\circ} C$ and $1$ atm is called its latent heat of vaporization. For water its value is $2367 k J k g^{- 1}$ .

Q. The change of energy on vaporizing 1.00kg of liquid water at 0∘C and 1atm is

2367kJkg−1

−2367kJkg−1

−2367kJmol−1

−2367kJg−1