Page 184 - The City and Guilds Textbook: Plumbing Book 1 for the Level 3 Apprenticeship (9189), Level 2 Technical Certificate (8202) and Level 2 Diploma (6035)

The City & Guilds Textbook: Plumbing Book 1

ACTIVITY Where:

Using the formula shown in L = litres
Example 1, calculate how Δt = temperature difference
many kilojoules it would
take to heat 140 litres of SHC of water = 4.186
water from 4°C to 65°C.
Therefore:
100 × (80−30) × 4.186 = 20930 kJ

Example 2
We can develop this concept further to calculate how many kilowatts it would
take to raise the temperature of the 100 litres of water by 50°C. To do this,
we need to state a time frame. Let us assume that the 100 litres of water is
required in one hour. The calculation would then become:

L × Δt × SHC of water
Time (in seconds)
Where:

ACTIVITY L = litres
Using the formula shown Δt = temperature difference
in Example 2, calculate SHC of water = 4.186
how many kilowatts it
would take to raise the 1 hour in seconds = 3600
temperature of the 140
litres of water from 4°C Therefore:
to 65°C in two hours. 100 × (80–30) × 4.186
= 5.81 kW
3600
Example 3
KEY POINT How many seconds would it take for 20 kg of water to be heated by 15°C using
Remember: water has a a 3 kW heating element?
specific heat capacity of
4.186 kJ/kg/°C and that The formula for this is:
1 W = 1 J/s. kg × t × SHC
kW
Where:

ACTIVITY kg = kilograms
Using the formula shown t = temperature
in Example 3, calculate kW = kilowatts
how many seconds it
would take for 42 kg SHC = specific heat capacity
of water to be heated
by 30°C using a 3 kW Therefore:
heating element. 20 × 15 × 4.186
= 418.6 s or 6.976 minutes
3

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