Question

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, and water vapor are 2.05 J/(g⋅°C), 4.18 J/(g⋅°C), and 2.01 J/(g⋅°C) respectively, the heat of fusion of ice is 6.01 kJ/mol, the heat of vaporization of water is 40.7 kJ/mol. (Please show your work)

a. Heating the H2) sample from –14°C to 58°C.

b. Cooling the H2O sample from 18°C to –2.3°C.

c. Cooling the sample from 140°C to 110°C.

d. Cooling the H2O sample from 97°C to 62°C.

e. Cooling the H2O sample from 102°C to 86.0°C

Answer 1

Solution :-

Lets assume we have 100 g water

Lets calculate the energy released in each case

a)heating water from -14 C to 58 c is the endothermic process therefore it does not release heat.

b) cooling sample from 18C to -2.3 C

q= (m*s*delta T)+(-m*delta Hfus)+(m*s*delta T)

=(100 g * 4.18J/gC*(0C-18C))+(-100g*334J/g)+(100g*2.05J/gC*(-2.3-0))

= -41154 J

c)cooling from 140 c to 110 C

q=m*s*delta T

= 100 g * 2.01J/gC*(110-140)

= -6030 J

d) cooling from 97 C to 62C

q=m*s*delta T

= 100 g * 4.18 J/gC * (62-97)

= -14630 J

e) cooling from 102 c to 86 C

q= (m*s*delta T) +(-m*delta Hvap)+(m*s*delta T)

   = (100g*2.01J/gC*(100-102))+(-100 g * 2259 J/g)+(100g*4.18J/gC*(86-100))

   = -232154 J

Therefore most of the energy is released in the case E