Calculate the temperature change when 500J of heat is supplied to 100g of water [specific heat capacity of water = 4200JKg$^{-1}$k$^{-1}$]

Analyze the question: We need the temperature rise (Δθ) when a known amount of heat is supplied to a known mass of water.

Concept: The heat absorbed by a substance is given by

Q = mcΔθ

so

Δθ = Q / (mc)

Solution:

• Q = 500 J
• m = 100 g = 0.1 kg (always convert to kg because c uses kg)
• c = 4200 J kg⁻¹ K⁻¹

Δθ = 500 / (0.1 × 4200) Δθ = 500 / 420 Δθ ≈ 1.19 °C ≈ 1.2 °C

Tip: Always convert grams to kilograms before using c = 4200 J kg⁻¹ K⁻¹. If you forget and use 100 g directly, you get 500/(100 × 4200) ≈ 0.001, which matches none of the options. Also, a change of 1 K equals a change of 1 °C, so the units are interchangeable for temperature differences.