A well-insulated electric water heater warms 125 kg of water from 20.0° C to 53.0° C in 29.0 min. Find the resistance of its heating element, which is connected across a 240-V potential difference.

Question

anonymous · Answer

Q=mc(T2-T1)

anonymous · Answer

R= (Delta V ^2 x time)/mc(T2-T1)

kropot72 · Answer

The energy input from the heating element = (wattage of element) * (heating time in seconds) = W * 29 * 60 joules
The specific heat of water remains close to 1.000 cal/(gm)(degree celsius) over the temperature range involved. The energy absorbed by the water = 125000 * (temperature rise) = 125000 * (53 - 20) = 125000 * 33 calories = 125000 * 33 * 0.239 joules
Equating the input energy with the absorbed energy:
$$W \times 29\times 60=125000\times33 \times0.239$$
Now can you solve for W?

ujjwal · Answer

$$\frac{V^2}{R}=\frac{ms\Delta \theta}{t}$$Both are relations for power. So, just equalize them. put the values.. and that's it...

anonymous · Answer

I already solved this lol.