Answer to Question #207990 in Classical Mechanics for Emma

Question

A 1kg of water initially at 25 degrees Celsius is brought to boil at 100 degrees. This process is then continued until all the water is completely turned into steam and evaporate. How much heat energy was used in the process if the Latent heat of vaporization is 2260kJ/kg?

Answer 1

Let's find the amount of heat required to convert 1 kg of water at 25°C to 1 kg of water at 100°C:

$$Q_1=m_wc_w\Delta T=1\ kg\cdot4180\ \dfrac{J}{kg\cdot\!^{\circ}C}\cdot75^{\circ}C=313500\ J.$$

Let's find the amount of heat required to convert 1 kg of water at 100°C to 1 kg of steam at 100°C:

$$Q_2=m_wL_v= 1\ kg\cdot2.260\cdot10^6\ \dfrac{J}{kg}=2260000\ J.$$

Finally, we can find the total amount of the heat required to convert 1 kg of water at 25°C to steam at 100°C:

$$Q=Q_1+Q_2=313500\ J+2260000\ J=2.57\cdot10^6\ J.$$