Цель: Понять
Решение:
Given:
- Resistance of the heating element, R = 110 Ω
- Voltage, V = 220 V
- Time, t = 5 minutes = 300 seconds
- Efficiency of the heater, η = 80%
- Volume of water, V = 0.5 liters
- Initial temperature of water, T_1 = 20°C
а) Чему равна мощность тока в электрокипятильнике?
To find the current flowing through the heating element, we can use Ohm's Law:
[ V = IR ]
Where:
- V is the voltage (220 V)
- I is the current
- R is the resistance of the heating element (110 Ω)
Solving for I:
[ I = \frac{V}{R} = \frac{220}{110} = 2 , A ]
So, the current flowing through the heating element is 2 Amperes.
б) Какое количество теплоты выделилось при работе электрокипя-тильника?
The power of the heating element can be calculated using the formula:
[ P = VI ]
Where:
- P is power
- V is voltage (220 V)
- I is current (2 A)
[ P = 220 \times 2 = 440 , W ]
The power of the heating element is 440 Watts.
The energy consumed by the heater in 5 minutes (or 300 seconds) can be calculated using the formula:
[ E = Pt ]
Where:
- E is the energy consumed
- P is power (440 W)
- t is time (300 s)
[ E = 440 \times 300 = 132,000 , J ]
Since the efficiency of the heater is 80%, the actual energy converted to heat is 80% of the above value:
[ Actual , Energy = 132,000 \times 0.8 = 105,600 , J ]
Therefore, the amount of heat produced by the heater is 105,600 Joules.
в) До какой температуры нагрелась вода?
The amount of heat produced by the heater is equal to the amount of heat absorbed by the water. The formula to calculate heat energy is:
[ Q = mcΔT ]
Where:
- Q is the heat energy
- m is the mass of water
- c is the specific heat capacity of water (assumed to be 4.18 J/g°C)
- ΔT is the change in temperature
Given that the volume of water is 0.5 liters = 500 grams (since the density of water is 1 g/cm³), we can calculate the change in temperature as follows:
[ Q = 105,600 , J ]
[ m = 500 , g ]
[ c = 4.18 , J/g°C ]
[ T_2 - T_1 = \frac{Q}{mc} = \frac{105,600}{500 \times 4.18} ≈ 50°C ]
Therefore, the water would heat up by approximately 50°C.
This completes the solution to the problem.
