How much heat (in joules) is required to warm 50.0 g of water from 20.0 °C to 80.0 °C? (Specific heat of water = 4.18 J/(g·°C))
A3,344 J
B8,360 J
C12,540 J
D250.8 J
Explanation
📌 q = m × c × ΔT, where q is heat, m is mass, c is specific heat, and ΔT is temperature change. q = 50.0 g × 4.18 J/(g·°C) × (80.0 − 20.0) °C = 50.0 × 4.18 × 60.0 = 12,540 J. Always compute ΔT as final minus initial; positive ΔT means heat absorbed, negative means heat released.
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