The electric iron marked with "220 V, 50 W" is connected to the 220 V power supply, and the heat generated per minute is______ J. One hour power consumption______ Degree

The electric iron marked with "220 V, 50 W" is connected to the 220 V power supply, and the heat generated per minute is______ J. One hour power consumption______ Degree


The electric iron is a pure resistance, ∵ P = 50W, t = 1min = 60s, ∵ the heat generated per minute is w = Pt = 50W × 60s = 3000j, and ∵ T1 = 1H = 3600s, ∵ the electric energy consumed for 1H is w = pT1 = 50W × 3600s = 180000j = 0.05kw · H = 0.05 degree



What should we pay attention to when measuring electric energy with electric energy meter and stopwatch


First of all, we should understand how to calculate the electric energy, q = Pt, P is the power, t is the time. P = UV, we deduce q = UVT, where u and V are measured with a multimeter (also can be measured directly with a power meter), t is measured with a stopwatch. Since it is measured with a meter, we first estimate the range, select the appropriate range, and the general measurement results of a digital meter are close to full deviation comparison



Resistance 55, voltage 220, working normally for 100s


Q=I²Rt=U²t/R=(220V)²*100s/55Ω=88000J



The name plate of an electric rice cooker is marked with "220V 1600W. If the actual voltage is 210v, what is the actual current in the electric rice cooker? What is the actual power? What is the power
The name plate of an electric rice cooker is marked "220V 1600W
(1) How much heat does this electric cooker produce when it works normally for 20 minutes?
(2) If this electric cooker works normally for 10 hours, how much power will it consume?
(3) What's the resistance of this rice cooker?
(4) If the actual voltage is 210v, what is the actual current and power in the rice cooker?


(1) Heat q = Pt = 1600W * 20min * 60s = 19200j (joule)
(2) Electric energy w = Pt = 1600W * 10h = 16000wh = 16kWh = 16kWh
(3) Resistance R = u ^ 2 / P = 220 V ^ 2 / 1600 w = 30.25 Ω
(4) Actual current I = u / r = 210v / 30.25 Ω = 6.94a
P=UI=210V*6.94A=1457.4W