It takes 56.6 joules of energy to raise the temperature of 150 milliliters of water from 5 degrees Celsius to 95 degrees 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 its final temperature?

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...

Answered by ollieboyne | 2024-06-10

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)

Answered by thomasgallagher98 | 2024-06-10

The electric heater will use approximately 0.0943 kilojoules of energy to heat the water to its final temperature. This is calculated based on the energy required (56.6 joules) and the heater's 60% efficiency. The final energy used is derived from the formula that takes into account the efficiency of the heater.
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Answered by ollieboyne | 2024-09-27