Question

How much heat is required to raise the temperature of 28.0 g of water from −35 °C to 145 °C?

Answer 1

heat require to raise temperature from -35 C to 0 C is given by,

Q1 = mC deltaT

where
Q1 = heat energy = ?
m = mass = 28 g
C = specific heat of ice = 2.06 J/g
ΔT = change in temperature = 35

Q1 = 28*2.06*35

Q1 = 2018.8 J

Heat required to raise the temperature of ice from -15 °C to 0 °C = 2018.8 J

Heat required to convert 0 °C ice to 0 °C water is given by,

Q2 = m·ΔHf

where

Q2 = heat energy = ?

m = mass = 28g

ΔHf = heat of fusion(334 J/g for water)

Q2 = 28*334

Q2 = 9352 J

Heat required to convert 0 °C ice to 0 °C water = 9352 J

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

Q3 = mcΔT

Q3 = 28*4.18*100

Q3 = 11704J

Heat required to raise the temperature of 0 °C water to 100 °C water = 11704 J

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

Q4 = m·ΔHvap
where
Q4 = heat energy
m = mass = 28g
ΔHvap= heat of vaporization = 2257 J/g

Q4 = 28*2257
Q4 = 63196 J

Heat required to convert 100 °C water to 100 °C steam = 63196J

Heat required to convert 100 °C steam to 145 °C steam

Q5 = mcΔT
Q5 = 28*2.09*45
Q5 = 2633.4 J

Heat required to convert 100 °C steam to 145 °C steam = 2633.4 J

Heat required Q = Q1+Q2+Q3+Q4+Q5

Q = 2018.8 + 9352 + 11704 + 63196 + 2633.4

Q = 88904.2 J

Heat required to convert -35 °C water to 145 °C is 88904.2 J or 88.9042 kJ