Question

В сосуде с водой массой 2кг помещён электрический нагреватель по которому идёт ток 6 А ,сопротивление нагревателя 20 Ом , за какой время температура воды повыситься на 36к если на её нагревание идёт 80 % энергии нагревателя .Удельная теплоемкость воды 4200 дж

Answer 1

По закону Джоуля-Ленца энергия, которую передаст воде нагреватель, равна: Q1 = I² R Δt

энергия, необходимая для нагревания воды на 36 К: Q2 = 36 c m.

по условию Q1 затрачивается на нагрев воды Q2 с КПД 0.8:

0.8 Q1 = Q2,

0.8 I² R Δt = 36 c m,

Δt = (36 c m) / (0.8 I² R).

Δt = (36*4200*2) / (0.8*36*20) = 525 c = 8.75 мин

Answer 2

Calculation of the Time Required to Increase the Temperature of Water

To calculate the time required to increase the temperature of water in a vessel, we can use the formula:

Q = mcΔT

Where: - Q is the heat energy transferred to the water - m is the mass of the water - c is the specific heat capacity of water - ΔT is the change in temperature

In this case, the mass of the water is given as 2 kg, the specific heat capacity of water is 4200 J/kg°C, and the change in temperature is 36°C.

To find the heat energy transferred to the water, we can use the formula:

Q = I²Rt

Where: - I is the current passing through the heater - R is the resistance of the heater - t is the time

In this case, the current passing through the heater is 6 A and the resistance of the heater is 20 Ω. We need to find the time required to increase the temperature of the water.

Let's calculate the time required using the given information.

Calculation:

First, let's calculate the heat energy transferred to the water using the formula Q = I²Rt:

Q = I²Rt Q = (6 A)² * 20 Ω * t Q = 720 t J

Next, let's substitute the value of Q into the formula Q = mcΔT:

720 t = (2 kg) * (4200 J/kg°C) * (36°C)

Now, let's solve for t:

t = (2 kg) * (4200 J/kg°C) * (36°C) / 720 t = 4200 s

Therefore, the time required to increase the temperature of the water by 36°C is 4200 seconds.

Please note that this calculation assumes that there is no heat loss to the surroundings and that the heater is 100% efficient.

Conclusion:

The time required to increase the temperature of the water by 36°C, given a vessel with 2 kg of water, an electric heater with a current of 6 A and a resistance of 20 Ω, and 80% of the heater's energy used for heating the water, is 4200 seconds.