Question

The heat capacity of water is 1cal degree'g1 (1 calorie per degree centigrade, per gram). You are given 1 gallon of water at 25 °C. Calculate the heat in Joules needed to boil this gallon of water under 1 bar pressure.

Answer 1

Answer:

The heat needed to boil 1 gallon of water is 81,490.62 Joules.

Explanation:

$Q=mc\Delta T$

Where:

Q = heat absorbed or heat lost

c = specific heat of substance

m = Mass of the substance  

ΔT = change in temperature of the substance

We have :

Volume of water = V = 1 gal = 4546.09 mL

Density of water , d= 1 g/mL

mass of water = m = d × V = 1g/mL × 4546.09 mL =  4546.09 g

Specific heat of water = c = 1 Cal/g°C

ΔT = 100°C - 25°C = 75 °C

9 (boiling pint of water is 100°C)

Heat absorbed by the water to make it boil:

$Q= 4546.09 g\times 1 Cal/g^oC\times 75^oC=340,956.75 Cal$

1 calorie = 4.184 J

$Q=\frac{340,956.75}{4.184} J = 81,490.62 J$

The heat needed to boil 1 gallon of water is 81,490.62 Joules.