How much heat is required to heat 300 g water from 20 degrees to boiling.
| January 31, 2021 | education
Given: m (mass of water for heating) = 300 g (0.3 kg); t (temperature of 300 g of water before heating) = 20 ºС.
Constants: С (specific heat capacity of water) = 4200 J / (kg * ºС); tboil (temperature of the beginning of the boiling process of water, n.o.) = 100 ºС.
The amount of heat required to bring 300 g of water to a boil is determined by the formula: Q = C * m * (tboil – t).
Calculation: Q = 4200 * 0.3 * (100 – 20) = 4200 * 0.3 * 80 = 100800 J.
Answer: Requires 100800 Joules of heat.
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