How much energy is required to heat 360 g h2o from a liquid at 65c to a gas at 115c


Introduction

In order to heat water from a liquid at 65 degrees Celsius to a gas at 115 degrees Celsius, a certain amount of energy is required. The amount of energy required will depend on the mass of water being heated and the temperature difference between the liquid and gas states.

The amount of energy required to heat water


The amount of energy required to heat water is dependent on the change in temperature of the water and the amount of water being heated. In general, it takes more energy to heat a larger quantity of water by a greater amount. For example, it takes more energy to heat 1 gallon (3.8 L) of water from 63 °F (17 °C) to boiling than it does to heat 1 pint (0.5 L) of water from 68 °F (20 °C) to 78 °F (26 °C).

The specific heat capacity of water is 4.184 J/g°C, which means that it takes 4,184 joules of energy to raise the temperature of 1 gram (g) of water by 1 degree Celsius (°C). To calculate the amount of energy required to raise the temperature of a given quantity of water by a given amount, you can use the following equation:

Q = m * c * ΔT

Where Q is the quantity of energy required (in joules), m is the mass of the water being heated (in grams), c is the specific heat capacity of water, and ΔT is the change in temperatureof the water (in degrees Celsius). Applying this equation to our previous example, we can calculate that it would take approximately 2,730 joulesof energy to heat 1 pint (0.5 L) of water from 68 °F (20 °C)to boiling:

Q = 0.5 L * 4,184 J/g°C * (100 °C – 20 °C)
Q = 2,730 J

The amount of energy required to vaporize water

The amount of energy required to vaporize water is called the latent heat of vaporization. It takes 2260 joules (J) of energy to turn 1 gram (g) of water from a liquid into a gas. So, it would take 823,600 J to vaporize 360 g of water—that’s a lot of energy!

Conclusion

In conclusion, it takes 84 kJ of energy to heat 360 g of water from 65ºC to 115ºC.


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