Help with some SEAI Second Level Math please

Calahonda52

From here
http://www.seai.ie/Schools/Post_Primary/Subjects/Physics/Unit_5_-_Heat_Quantity/Sample_Problems_-_Specific_Heat_Capacities/

Comes the following: I am puzzled by the 103 in question 2

It does not impact the answer as it is in both sides of the heat equation

Exam tip:
When solving problems involving specific heat capacities, you only need to know two equations!

1. Heat = mass × specific heat capacity × temperature change
Q = mc Dq
and

2. Heat lost = Heat gained
mc Dq = mc Dq

When reading the question, concentrate on what is losing heat and what is gaining heat - then use the above two equations to solve the problem.

Worked Examples
Question 1

Given that the specific heat capacity of water is 11 times that of copper, calculate the mass of copper at a temperature of 100 °C required to raise the temperature of 200 g of water from 20.0 °C to 24.0 °C, assuming no energy is lost to the surroundings.

Solution

Heat lost by copper = heat gained by water
mcuccuDqcu = mwcwDqw
mcuccu(100 - 24) = 0.200 × 11ccu(24 – 20)
76mcu = 8.8
mcu = 0.116 kg

Question 2
Three litres of water at 100 °C are added to 15 litres of water at 40 °C. Calculate the temperature of the mixture. Take the mass of 1 litre of water to be 1 kg and the specific heat capacity of water to be 4.2 × 103 J kg -1 K -1

Solution

Let the temperature of the mixture q
Heat lost = heat gained
m1c1Dq1 = m2 c2 Dq2
3 × 4.2 × 103 × (100 – q) = 15 × 4.2 × 103 × (q -40)
Solving for q gives q = 50°C

Man of Aran

Likely it's a typo, should be 10 to power3 to get convert 1kg mass to grams since SHC has units [ joule / gram per deg C]. If I understand your query correctly.

Calahonda52

Good spot, many thanks indeed.