Question

How much thermal energy (in kcal) is required to change a 46 g ice cube from a solid at - 6.6 ^oC to steam at 21.1 ^oC above boiling?

Answer 1

Given values are, mass of ice = 46 g, temperature of ice = -6.6 = -266.4 K, temperature of steam = 121.1 = 394.1 k

Values to be used:

specific heat of ice = 2.06 J/g-K, specific heat of water = 4.18J/g-K, specific heat of steam = 1.87J/g-K, latent heat of fusion = 334J/g, latent heat of vaporization = 2258J/g

To calculate the thermal energy required to change ice in to steam, we have to consider following step

1) Energy required to convert -6.6 ice to 0 ice = ms = 46*2.06*6.6 = 94.76J

2) Energy required to convert 0 ice to 0 water = mL = 46*334=15364J

3) Energy required to convert 0 water to 100 water = ms = 4.18*46*100 = 19228J

4) Energy required to convert 100 water to 100 steam = mL = 46*2258 = 103868J

5)Energy required to convert 100 steam to 121.1 steam = ms = 46*1.87*21.1 = 1815J

Hence total thermal energy required will be equal to the sum of all energy calculated above and is = 140369.76J

In kCal units = 160369.76/4.18*10^3 = 33.54kCal