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Calculate the heat needed to change 3.00 * 10^2 grams of water at 25 degrees celsius to steam at 100 degrees celsius

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Answer:

9.4×10⁴J

Step-by-step explanation:

The specific heat capacity (SHC) of water, which is the amount of energy (J) that it takes to warm 1mL (1mL = 1g) of water by 1°C is 4.184 J/°C.

Q = mcΔT relates energy to mass, SHC, and change in temperature.

The mass of water is 300g, and will not change because no water will evaporate until it reaches 100°C.

The change in temperature is 75°C:

100°C - 25°C = 75°C

(Desired temp. - initial temp. = change in temp.)

Subbing these values into Q=mcΔT:

Q = 300g × 4.184J/g×°C × 75°C

Q = 94140J

Scientific notation tells us that we have two significant figures, so:

Q = 9.4 × 10⁴J

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