Question

Каким количеством теплоты можно нагреть 0,3 кг воды от 12 градусов цельсию до 20градусов цельсию

Answer 1

Q=m*c(t1-t0)
C(воды)=4200Дж/кг
Q=0.3*4200*(20-12)
Q=1260*8=10080Дж=10.08кДж

Answer 2

Calculating Heat Energy Required to Heat Water

To calculate the heat energy required to heat 0.3 kg of water from 12 degrees Celsius to 20 degrees Celsius, we can use the formula:

Q = mcΔT

Where: - Q = heat energy - m = mass of the substance (in kg) - c = specific heat capacity of the substance (in J/kg°C) - ΔT = change in temperature (in °C)

Given: - m = 0.3 kg - Initial temperature (t1) = 12°C - Final temperature (t2) = 20°C

Using the specific heat capacity of water, which is approximately 4200 J/kg°C, we can calculate the heat energy required.

Q = mcΔT = 0.3 kg * 4200 J/kg°C * (20°C - 12°C)

Q = 0.3 * 4200 * 8 = 10080 J

So, the heat energy required to heat 0.3 kg of water from 12°C to 20°C is 10080 Joules.

Cited Sources

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