Question

Hola mate. ☹ ☹ ☹

✒✒✒✒✒✒✒✒✒✒✒


time required required to boil 2 litre of water initially at 20°C by heater coil which works at a 80% efficiency spending 500 j/S is??

No spam...
✴✴✴✴✴✴✴✴✴✴​

Answer 1

The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C.

A heater coil at 80% efficiency would add 400 joule/s to the water.

Raising 2000 grams of water 80 °C requires 2000*4.186*80 joules, or 669,760 joules.

At 400 joule/s, delivering 669,760 joules would take 1,339.52 seconds - a little over 22 minutes - if we ignore any other heat losses.

Answer 2

The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C.

A heater coil at 80% efficiency would add 400 joule/s to the water.

Raising 2000 grams of water 80 °C requires 2000*4.186*80 joules, or 669,760 joules.

At 400 joule/s, delivering 669,760 joules would take 1,339.52 seconds - a little over 22 minutes - if we ignore any other heat losses.