Räkna ut temperaturen av en bunsen-låga

Jag hoppas att det går att förstå frågan även om jag bara postar svaret på frågan. 
Det är helt enkelt så att jag inte förstår sista delen av uträkningen, jag fetstilar den delen.
Varför behövs den? Vad skulle hända om den fetstilade delen inte skulle vara med?

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After being placed into the calorimeter the heat
energy lost by the piece of brass is transferred
into the water as well as the aluminium walls of
the calorimeter, which are in contact with the
water. We therefore need to calculate the heat
transferred to the water and to the aluminium
calorimeter using the appropriate specific heat
capacities.
Temperature change for water and aluminium
calorimeter (ΔT) = (77.50 – 24.50) °C = 53.00 °C
= 53.00 K
Heat change (water) = mcΔT
= 200.00 g × 4.18 J g–1 K–1 × 53.00 K
= 4.43 × 104 J
Heat change (aluminium) = mcΔT
= 80.00 g × 0.900 J g–1 K–1 × 53.00 K
= 3.82 × 103 J
Total heat energy gained by water and aluminium
= (4.43 × 104 J) + (3.82 × 103 J) = 4.81 × 104 J
Therefore heat energy lost by the brass = 4.81 ×
104 J
Heat change (brass) = mcΔT
4.81 × 104 J = 21.20 g × 0.400 J g–1 K–1 × ΔT
ΔT = 5670 K (5670 °C)
Initial temperature = Tf + ΔT = 77.50 °C +
5670 °C = 5748 °C
The temperature of the flame is 5748 °C.