Specific Heat Capacity

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Take two beakers, one filled with oil and one filled with water. Place both on an identical heat source, for the same amount of time. You predict that they will both be raised to the same temperature, but you would be incorrect. Different substances will heat differently to other substances.  This is related to a concept called specific heat capacity, the idea of how much a substance temperature might rise when a certain amount of energy is applied to it.

This video examines the physics principles,  includes some everyday examples, and also provides some mathematical modelling.
Also examine the worked example, and try some problems yourself.
Test your understanding
Interactive
The following interactive is a great way to simulate qualitatively, the idea of specific heat capacity. Click INTRO
  1. Start by placing both the beaker of water and oil on the bunsen burner
  2. Click both energy symbols and link heaters
  3. Add a thermometer to both containers by dragging
  4. Start heating both liquids.
  5. You will note that the water will rise in temperature at a SLOWER rate than oil, as it has a greater heat capacity 
  6. See if you can work out the order of the specific heat capacity of the four material provided
Sample Problem
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
Picture

Some problems to try

(Note: Specific heat capacity of water c = 4182 J/kg°C)

1. How much energy must be provided to raise the temperature of 400 mL of water by 5°C?

Given: Volume V = 400 mL → Mass m = 0.400 kg, ΔT = 5°C, c = 4182 J/kg°C

Formula: q = m × c × ΔT

Calculation:

q = 0.400 kg × 4182 J/kg°C × 5°C

q = 8364 J

Answer: 8364 J (or 8.36 kJ)
2. 5 kJ of heat is applied to 125 mL of water initially at 20°C. What is its new final temperature?

Given: Heat q = 5000 J, Volume V = 125 mL → Mass m = 0.125 kg, Tinitial = 20°C

Formula: ΔT = q / (m × c)

Calculation:

ΔT = 5000 / (0.125 × 4182) = 5000 / 522.75 ≈ 9.56°C

Tfinal = Tinitial + ΔT = 20°C + 9.56°C = 29.56°C

Answer: 29.56°C
3. An unknown 250 g metal block is heated to 100°C and dropped into 200 mL of water at 20°C. If the final equilibrium temperature of the mixture is 28°C, what is the specific heat capacity of the metal? (Assume no heat is lost to the surroundings)

Given: Water mass mw = 0.200 kg, Metal mass mm = 0.250 kg

ΔTw = 28°C - 20°C = 8°C

ΔTm = 100°C - 28°C = 72°C

Heat balance: Heat gained by water = Heat lost by metal

mw × cw × ΔTw = mm × cm × ΔTm

0.200 × 4182 × 8 = 0.250 × cm × 72

6691.2 = 18 × cm

cm = 6691.2 / 18 ≈ 371.7 J/kg°C

Answer: 371.7 J/kg°C (approx. Copper)
4. A 1500 W electric kettle contains 1.2 L of water at 18°C. Assuming 85% heating efficiency, how long will it take to bring the water to its boiling point (100°C)?

Given: Power P = 1500 W, Mass m = 1.2 kg, Efficiency = 85% (0.85)

ΔT = 100°C - 18°C = 82°C

Energy required: q = m × c × ΔT = 1.2 × 4182 × 82 = 411,508.8 J

Effective power input: Peffective = 1500 W × 0.85 = 1275 W (J/s)

Time required: t = q / Peffective = 411,508.8 / 1275 ≈ 322.75 seconds

Answer: 323 seconds (approx. 5 minutes 23 seconds)
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