A 20.0 g sample of aluminum (specific heat = 0.902) g-1 oC-1) with an initial temperature of 2.5°C is heated with 427 J of energy. What is the final temperature of the sample?
O 72.3°C
O 23.7°C
0 26.2°C
0 74.8°C
O 24.9°C

T(final) = 26.2°C (3 sig. figs.)

Explanation:

Q = m·c·ΔT

m = mass of sample of interest = 20.0g

c = specific heat of sample of interest = 0.902j·g⁻¹·C⁻¹

ΔT = Temperature change = T(final) - T(initial) = T(f) - 2.5°C

Q = 427 joules

Q = m·c·ΔT => = ΔT = Q/m·c => T(f) = Q/m·c + T(i) = (427j/20.0g·0.902j·g⁻¹·C⁻¹) + 2.5°C = 26.16962306°C (calc. ans.) ≅ 26.2°C (3 sig. figs.)