Heating Curve of Water
SECTION (A): Solid State
 temperature increase
 specific heat of solid
 energy increases
SECTION (D): Condensation/Evaporation
 temperature stays the same
 latent heat of vaporization
 energy increases
SECTION (B): Freezing/Melting
 temperature stays the same
 latent heat of fusion
 energy increases
SECTION (E): Gas State
 temperature increases
 specific heat of gas
 energy increases
SECTION (C): Liquid State
 temperature increases
 specific heat of liquid
 energy increases
Calculating Heat during a Temperature Change
 Heat (or energy) can be measured in units of calories or Joules.
 When there is a temperature change (∆T), heat (Q) can be calculated using this formula:
Q = mC∆T
Q = heat (Joules)
m = mass (g)
Cp = specific heat
∆T = change in temperature (T2 – T1 in ⁰C)
Calculating Heat during a Phase Change
 During a phase change, there is no change in temperature. So, the heat used to change the phase of matter
can be calculated using this formula:
Q = m(Hf or Hv)
Q = heat (Joules)
m = mass (g)
Hf = latent heat of fusion
Hv = latent heat of vaporization
Energy Calculation Constants for Water
Specific Heat (Cp) of water = 4.18 J/g ⁰C
Specific Heat (Cp) of steam = 2.02 J/g ⁰C
Specific Heat (Cp) of ice = 2.05 J/g ⁰C
Latent Heat of Fusion for H2O = 334 J/g
Latent Heat of Vaporization for H2O = 2260 J/g
NOTE: Energy calculations are at a pressure of 1 atm and is a different value at any other pressure
Heating Curve of Water
Example Problems for H2O
How many Joules of heat are needed to change the temperature of 25 grams of steam from 105⁰C to 135⁰C?
Temperature change: Q = mC∆T
Q = (25)(2.02 J/g ⁰C)( 135⁰C - 105⁰C)
Q = 1,515 Joules
How many Joules of heat are required to melt 35 grams of ice at 0⁰C?
Phase Change (Melting Point): Q = mHf
Q = (35)(334 J/g)
Q = 11,690 Joules
How many Joules of heat energy are required to change a 13.5 gram ice cube at -11.4⁰C to liquid water at
45.9⁰C?
Temperature change (Initial Temperature to Melting Point): Q = mC∆T
Q = (13.5)(2.05 J/g ⁰C)(0⁰C – [-11.4⁰C])
Q = 315.495 Joules
Phase Change (Melting Point): Q = mHf
Q = (13.5)(334 J/g)
Q = 4,509 Joules
Temperature change (Melting point to Final Temperature): Q = mC∆T
Q = (13.5)(4.18 J/g ⁰C)(45.9⁰C - 0⁰C)
Q = 2,590.137 Joules
TOTAL REQUIRED: 315.5 Joules + 4,509 Joules + 2,590 Joules = 7,415 Joules
To answer these types of problems you need to first identify if the problem is dealing with a temperature change (
part A, C and E of the curve) or a phase change ( part B and D). Once you know what type of problem it is, you
write down the proper equation(Q = mC∆T for temperature change, Q = mH for phase change).
If the problem is dealing with a temperature change, you need to know what number to plug in for C. You use 2.05
if it is ice, 4.18 if it is water and 2.02 if it is steam. C represents heat capacity/specific heat, which is how much
energy is required to change the temperature of a gram of substance by a degree. Water takes more energy to change
than ice and steam, that’s why you use a different number for each state.
If the problem is dealing with a phase change, you need to know what number to plug in for H. H represents latent
heat, which is how much energy is required to change the state of a gram of substance. H = 334 for the transition
between ice and water. H = 2260 for the transition between water and steam.
Some of the problems require you to rearrange the equation. You rearrange the equation for the variable that the
question wants you to solve for and plug in everything else from the problem.
Heating Curve of Water
Additional Problems for H2O
1.) How many Joules of heat are given off when 5 grams of water cool from 75⁰C to 25⁰C?
Temperature Change: Q = mC∆T
Q = (5)(4.18)(50)
Q = 1045J
2.) If 575 Joules of heat are added to 45 grams of water at 25⁰C, what will the new temperature be?
Temperature Change: Q = mC∆T
Q
𝑚𝐶
(575)
∆T =
(45)(4.18)
∆T = 3.05
T1 + ∆T = T2
25 + 3.05 = 28.05⁰C
∆T =
3.) How many Joules does it take to boil 35 g of water at 100⁰C?
Phase Change: Q = mHv
Q = (35)(2260)
Q = 79,100
4.) When 1550 Joules of heat are added to a 30.7 gram sample of H2O, the temperature changes by
25⁰C. Is the H2O in solid, liquid, or gas form?
Temperature Change: Q = mC∆T
Q
𝑚∆T
1150
C=
(30.7)(25)
C = 2.02 (Steam)
C=
5.) How many Joules of heat are necessary to raise the temperature of 25 grams of steam from
110⁰C to 160⁰C?
Temperature Change: Q = mC∆T
Q = (25)(2.02)(50)
Q = 2525J
6.) How many Joules are given off when 55 grams of water at 0⁰C freezes?
Phase Change: Q = mHf
Q = (55)(334)
Q = 18,370J
7.) In a lab experiment, students were given a sample of water at 22.5⁰C. The students added 935
Joules of heat to the sample, and the temperature increased to 49.2⁰C. What is the mass of the
sample of water?
Temperature Change: Q = mC∆T
Heating Curve of Water
m=
Q
𝑐∆T
935
(4.18)(49.2 − 22.5)
m = 35g
m=
8.) If 4,550 Joules of heat are added to 17.5 grams of water, by how many degrees Celsius would
the water increase?
Temperature Change: Q = mC∆T
∆T =
Q
𝑚C
4550
(17.5)(4.18)
∆T = 62̊C
∆T =
9.) How many Joules of heat are required to change 25 grams of water at 83.2⁰C to steam at 100⁰C?
PART 1: Temperature Change: Q = mC∆T
Q = (25)(4.18)(100-83.2)
Q = 1755.6J
PART 2: Phase Change: Q = mH
Q = (25)(2260)
Q = 56500J
FINAL ANSWER: 1,755.6J + 56,500J = 58,255.6J
10.) How much heat (in Joules) is required to change a 23 gram ice cube from -12.7⁰C to liquid water
at 19.4⁰C?
PART 1: Temperature Change: Q = mC∆T
Q = (23)(2.05)(12.7)
Q = 598.8J
PART 2: Phase Change: Q = mH
Q = (23)(334)
Q = 7682J
PART 3: Temperature Change: Q = mC∆T
Q = (23)(4.18)(19.4)
Q = 1865.1J
FINAL ANSWER: 598.8J + 7682J + 1865.1J = 10145.9J