JAMB - Physics (2012 - No. 17)
Calculate the temperature change when 500J of heat is supplied to 100g of water
[specific heat capacity of water = 4200JKg\(^{-1}\)k\(^{-1}\)]
12.1oC
2.1oC
1.2oC
0.1oC
Explicação
\(\theta = \frac{Q}{MC}\)
\(\theta = \frac{500}{0.1 \times 4200}\) ≈ 1.2 ºC
\(\theta = \frac{500}{0.1 \times 4200}\) ≈ 1.2 ºC