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How much heat, in calories, does it take to warm 890 g of water from 12.0 ∘C to 45.0 ∘C?
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- Anonymous2 years agoFavorite Answer
calorie is defined as what raises the temperature of a gram of water by one degree Celsius, so 890 times 33
- 2 years ago
Q = m * c * dT
dT = 45 - 12 = 33 degrees Celsius
m = 890 grams
c = 4.186 Joules per degree change in Kelvin per gram
890 * 4.186 * 33 Joules =>
122942.82
There are 4.186 Joules per Calorie, so we can actually make this nicer
890 * 33 =>
89 * 11 * 3 * 10 =>
267 * 11 * 10 =>
2937 * 10 =>
29370 calories
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