Question

What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization of water is 2.25 x 10^6 J kg-1.

Answer 1

Answer:

$\boxed{\text{889 g}}$

Explanation:

The formula relating the mass m of a sample and the heat q to vaporize it is

q = mL, where L is the latent heat of vaporization.

$\begin{array}{rcl}2000 \times 10^{3} \text{ J} & = & m \times \dfrac{2.25 \times 10^{6} \text{ J}}{\text{1 kg}}\\\\m & = & \dfrac{2000 \times \times 10^{3}\text{ kg}}{2.25 \times 10^{6}}\\ & = & \text{0.889 kg}\\\\ & = & \text{889 g}\\\end{array}\\\text{The mass of water is \boxed{\textbf{889 g}} }$