Question

Решите задачу по физике 8 класс PLIZ Какое количество теплоты требуется для нагривания 20 кг воды от 18 до 28С

Answer 1

Дано:
Дано:
m = 20 кг
t1 = 18 градусов
t2= 28 градусов
C = 4200 дж/кг * с
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Q - ?
   
                                             Решение
Q =CmΔt                 
t = t2 - t1
Q = 4200 * 20 * 10 = 840000 Дж =840кДж

Answer 2

Calculating the Heat Required to Raise the Temperature of Water

To calculate the amount of heat required to raise the temperature of water, we can use the formula:

Q = mcΔT

Where: - Q = heat energy (in joules) - m = mass of the substance (in kilograms) - c = specific heat capacity of the substance (in joules per kilogram per degree Celsius) - ΔT = change in temperature (in degrees Celsius)

Given: - m = 20 kg (mass of water) - ΔT = 28°C - 18°C = 10°C (change in temperature) - The specific heat capacity of water is approximately 4.18 J/g°C or 4186 J/kg°C.

Using the formula, we can calculate the heat required to raise the temperature of the water.

Calculation

Q = mcΔT Q = 20 kg * 4186 J/kg°C * 10°C Q = 836,000 J

So, the amount of heat required to raise the temperature of 20 kg of water from 18°C to 28°C is 836,000 joules.