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it takes 56.6 joules of energy to raise the temperature of 150 milliliters of water from 5 degrees Celsius to 95 degree Celsius if you use an electric water heater that is 60% efficient how many kilojoules of energy will the heater actually use by the time the water reaches it's final temperature

Sagot :

ollieboyne
If 56.6J = 60%,

then 56.6/0.6 =  94.3333 J = 0.0943... kJ

The rest of the information (mL and temperature change) isn't needed for the answer...
thomasgallagher98
ANSWER: 94.33 J
EXPLANATION
I'm assuming this is a multi-part question as most of these values aren't needed for this specific question as if we know that 56.6 J is needed but the heater only uses 60% efficiently, 56.6 is 60% of the heater's total required value to raise this amount of water by this much.
Simply multiply up:
60% = 56.6
⇒ 10% = 9.4333
⇒ 100% = 94.33333.... J
(although we cannot have recurring decimals in Physics so stop at a certain point and mention the number of decimal places/significant figures)

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