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q = m x C x ΔT

q = amount of heat energy gained or lost by substance in Joules

m = mass of sample in grams = 20g

C = heat capacity (J/g•oC) = 4.184 J/g•oC

Tf = final temperature = 55 oC

Ti = initial temperature = 5 oC

ΔT = (Tf - Ti) = (55 oC - 5 oC) = 50 oC

q = 20g x 4.184J/g•oC x 50 oC = 4184 Joules

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11y ago
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10y ago

Nominally 700,000 calories, or about 2,928,800 joules.

(That's a lot ! It's enough energy to lift a ton of gravel

more than 1,000 feet straight up off the ground.)

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8y ago

The heat capacity of water is 4184 J/kg/deg, so to raise 0.25 kg by 10 degrees (20-30 deg C), it would take (4184 J/kg/deg)(0.25 kg)(10 deg) = 10,460 J = 10.46 kJ = 10. kJ (2 sig figs)

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8y ago

The necessary heat is 2,5 kcal.

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7y ago

Approx 10.4 kilojoules.

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6y ago

The necessary heat is 9,42 kJ.

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Q: How much heat is required to raise the temperature of 0.25 kg of water from 20C to 30C?
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