Calculate the temperature change when 500J of heat is supplied to 100g of water
[specific heat capacity of water = 4200JKg\(^{-1}\)k\(^{-1}\)]
\(\theta = \frac{Q}{MC}\)
\(\theta = \frac{500}{0.1 \times 4200}\) ≈ 1.2 ºC
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