
Воду массой 1,5 кг нагрели до температуры кипения за 5 мин.Мощность электрического чайника 2
кВт.КПД чайника-84%.Какова была начальная температура воды? Помогите пожалуйстааа ооооочень нужно срочно.Буду очень благодарна!!! Заранее спасибо!

Ответы на вопрос

m=1.5 кг τ=5*60=300 с P=2000 Вт c=4200 η=0,84 t2=100 C t1=?
η=Q/(P*τ) Q=c*m*(T2-t1)
t1=t2-η*P*τ/(c*m)=100-0.84*2000*300/(4200*1.5)=20 С
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To find the initial temperature of the water, we can use the formula:
Q = m * c * ΔT
Where: - Q is the heat energy transferred to the water (in Joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (approximately 4.18 J/g°C or 4180 J/kg°C) - ΔT is the change in temperature (final temperature - initial temperature)
First, let's convert the mass of the water from kilograms to grams: - 1.5 kg = 1500 g
Next, let's calculate the heat energy transferred to the water: - Q = m * c * ΔT
We know that the power of the electric kettle is 2 kW (kilowatts) and the efficiency (η) is 84%. The power is given by the formula:
P = Q / t
Where: - P is the power (in watts) - Q is the heat energy transferred to the water (in Joules) - t is the time taken to heat the water (in seconds)
We can rearrange the formula to solve for Q:
Q = P * t
Given that the power is 2 kW and the time is 5 minutes (300 seconds), we can calculate Q:
Q = 2 kW * 300 s = 600 kJ = 600,000 J
Now, let's substitute the values into the first formula and solve for the initial temperature (T1):
Q = m * c * ΔT 600,000 J = 1500 g * 4.18 J/g°C * (100°C - T1)
Simplifying the equation:
600,000 J = 6270 g°C * (100°C - T1) 600,000 J = 627,000 g°C - 6270 g°C * T1 6270 g°C * T1 = 627,000 g°C - 600,000 J 6270 g°C * T1 = 27,000 g°C T1 = 27,000 g°C / 6270 g°C T1 ≈ 4.31°C
Therefore, the initial temperature of the water was approximately 4.31°C.
Please note that this calculation assumes that the specific heat capacity of water remains constant over the temperature range and that there are no heat losses to the surroundings.


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