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How much heat energy must be absorbed by 300 g of liquid water to raise its temperature by 30 degree straight C?

How much heat energy must be absorbed by 300 g of liquid water to raise its temperature-example-1
User Systho
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Answer: To calculate the heat energy required to raise the temperature of a substance, we use the formula:

Q = m * c * ΔT

where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

For liquid water, the specific heat capacity is approximately 4.18 J/g°C.

So, for 300 g of liquid water to be raised by 30°C, we have:

Q = 300 g * 4.18 J/g°C * 30°C

Q = 37,620 J

Therefore, 37,620 Joules of heat energy must be absorbed by 300 g of liquid water to raise its temperature by 30 degrees Celsius.

User Sam Grondahl
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