Question

2.09 J>g °C, and that of steam is 2.01 J>g °C. 72. How much heat (in kJ) is evolved in converting 1.00 mol of at - 10.0 °C, to steam at 110.0 °C? The heat capacity of ice is ## 2.01 J>g °C, and that of ice is 2.09 J>g °C. Phase Diagrams steam at 145 °C to ice at - 50 °C? The heat capacity of steam is ##

Answer 1

1,54.95KJ

2. phase diagram can be found as attached

Step-by-step explanation:

Definition of terms

Heat is the degree of hotness or coldness in a body

Specific heat capacity is the amount of heat to raise one kg mass of a substance by 1 degree rise in temperature

How much heat (in kJ) is evolved in converting 1.00 mol of iceat - 10.0 °C, to steam at 110.0 °C?

we take it step by step

the heat needed to raise ice from -10 to 0C

Q=mcdT

mass=mole*relative molecular mass of water

mass (g)=1*18g/mol

mass=18g

Q=18*2.1*(0-(-10)=378J

2. the heat of fusion of ice

Qf=mlf

Qf=18*334J/g

Qf=6012J

3. heat to take water from 0c to 100c

Q=18*4.18*(100)

7524J

4. heat of vapourization

Qv=mLv

Qv=18*2260=40680J

5. heat to raise the steam from 0c to 110c

Q=mCsteam*dT

Q=18*2.01*(110-100)

Q=361.8J

add up all the heat evolved

378+6012+7524+40680+361.8

=54955.8

54.95KJ