Question

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:

A. joules needed to melt 90.0 g of ice at 0 ∘C and to warm the liquid to 85.0

B. kilocalories released when 30.0 g of steam condenses at 100 ∘C and the liquid cools to 0 ∘C

C. kilojoules needed to melt 26.0 g of ice at 0 ∘C, warm the liquid to 100 ∘Cand change it to steam at 100 ∘C

∘C

Answer 1

(a)

Heat needed to melt ice = m*L = 90*334 = 30060 J

Heat needed to warm water = m*C*dT = 90*4.184*(85-0) = 32007.6 J

So,

Total heat needed = 30060+32007.6 = 62067.6 J

(b)

Heat released by cooling of steam = m*L = 30*2230 = 66900 J

Heat released by cooling of water = m*C*dT= 30*4.184*(100-0) = 12552 J

So,

Total heat released = 66900+12552 = 79452 J = 18.98 kcal

(c)

Heat needed to melt ice = m*L = 26*334 = 8684 J

Heat needed to warm water = m*C*dT = 26*4.184*(100-0) = 10878.4 J

Heat needed to vaporise water = m*L = 26*2230 = 57980 J

So,

Total heat needed = 8684+10878.4+57980 = 77542.4 J = 77.54 kJ

Hope this helps !