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How many Joules of heat are required to raise the temperature of 20.0 grams of water from 30.0oC to 40.0oC?

Sagot :

Timo86idn

Answer:

840 J

Explanation:

c ≈ 4200 J / (kg * °C)

m = 20 g = 0,02 kg

[tex]t_{1}[/tex] = 30 °C

[tex]t_{2}[/tex] = 40 °C

The formula is: Q = c * m * ([tex]t_{2} - t_{1}[/tex])

Calculating:

Q = 4200 * 0,02 * (40 - 30) = 840 (J)

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