Question

calculate the amount of heat which is required to convert 25 gram of ice at minus 15 degree Celsius into steam at hundred degree Celsius

Answer 1

I am using some standard values
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C


Ur solution

Heat required to raise the temperature - 15 °c to 0°c

Q=mcΔT

where
q = heat energy
m = mass
c = specific heat

ΔT = change in temperature

Q= (25 g)x(2.09 J/g·°C)[(0 °C - 15°C)]
q = (25 g)x(2.09 J/g·°C)x(15 °C)
q = 783.75J

Heat required to raise the temperature of ice from -15 °C to 0 °C = 783.75 J


====>Heat required to convert 0 °C ice to 0 °C water

q = m·ΔHf

where
q = heat energy

m = mass

ΔHf = heat of fusion(334 j/g for water)
q = (25 g)x(334 J/g)
q = 8350 J

Heat required to convert 0 °C ice to 0 °C water = 8350 J


Heat required to raise the temperature of 0 °C water to 100 °C water

q = mcΔT

q = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]
q = (25 g)x(4.18 J/g·°C)x(100 °C)
q = 10450 J

Heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J

Heat required to convert 100 °C water to 100 °C steam

q = m·ΔHv

where
q = heat energy
m = mass
ΔHv = heat of vaporization

q = (25 g)x(2257 J/g)
q = 56425 J


total heat energy

HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4

HeatTotal = 783.75J + 8350 J + 10450 J + 56425 J 

HeatTotal = 76008.75J



This is ur ans hope it will help you in case of any doubt comment below

Answer 2

dH = msdt+mL+msdt+mL

dH = 25×0.5×(0 degree+(-15 degree) )

+ 25×80 + 25×1×100 degree +

25×540

dH = 25×0.5×15 + 2000 + 2500 +

13500

dH = 187.5 + 2000 +2500+13500

dH = 187.5 + 18000

dH = 18187.5 cal