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The specific heat of iron is either 0.46 or 0.45 Jolules/grams*Celsius, so.........

q(in Joules) = mass * specific heat * Temp final - Temp. initial

q = (65 grams Fe)(0.46 J/gC)(95 C - 25 C)

= 2093 joules of energy

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13y ago
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13y ago

Specific heat of iron is 0.46 J/gC so, use;

q = mass * specific heat * change in temp.

q = (35 g Fe)(0.46 J/gC)(35 C - 25 C)

= 161 Joules (1 calorie/4.184 Joules)

= 38 calories

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7y ago

The formula is: 0,108 x 3000 x (T1 -T2), in kilocalories.

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7y ago

heat = 3000 g x 0.450 J/g/deg x 5 deg = 6,750 Joules or 6.75 kJoules

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Q: How many joules of energy are required to raise the temperature of 65 grams of iron from 25 degrees celsius to 95 degrees celcsius?
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