Question

How much heat (in KJ) is required to warm 13.0 g of ice, initially at -10.0 *C, to steam at 113.0 degress C? The heat capacity of ice is 2.09 J/g degress C and that of steam is 2.01 J/g * C. Show work.

Answer 1

Q = heat change for conversion of ice at -10 ^oC to ice at 0 ^oC + heat change for

conversion of ice at 0^oC to water at 0^oC + heat change for conversion of water at 0^oC to water at 100 ^oC +heat change for conversion of water at 100 ^oC to vapour at 100 ^oC+ heat change for conversion of vapour at 100 ^oC to vapour at 113 ^oC

Amount of heat required , Q = mcdt + mL + mc'dt + mL' + mc"dt"

                                              = m(cdt + L + c'dt' + L' + c"dt" )

Where

m = mass of ice = 13.0 g

c'' = Specific heat of steam = 2.01 J/g degree C

c' = Specific heat of water = 4.186 J/g degree C

c = Specific heat of ice= 2.09 J/g degree C

L’ = Heat of Vaporization of water = 2260 J/g

L= Heat of fusion of ice = 334.9 J/g

dt’’ = 113-100 = 13^oC

dt' = 100 -0 =100 ^oC

dt = 0-(-10)=10 ^oC

Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" )

                                      = 39.8 x10³ J

                                     = 39.8 kJ

Therefore the amount of heat required is 39.8 kJ