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Water has a specific heat of 1 calorie per gram per degree. This means that in order to raise 28.4 grams of water by 1 degree Celsius, it takes 28.4 calories. At an increase of 44.6 degrees Celsius, this equates to 1,266.64 calories of heat.

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9y ago
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13y ago

q(Joules) = mass * specific heat * change in temp.

q = 32.0 grams H2O * 4.180 J/gC *n (80.0 C - 25.0 C)

= 7230.08 Joules

7230.08 Joules (1 calorie/4.184 Joules)

= 1728 calories

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12y ago

The total amount of heat is the same as the energy. Because no phase change is occurring, use the formula Q = mc(delta)T. Q is the energy, m is the mass, c is the specific heat capacity of water as a liquid, and (delta)T is the change in temperature. So: Q = (100.0 g)(4.186 J/g°C)(15°C); Q = 6279 Joules.

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10y ago

I work in joules, so a conversion at the end.

q(joules) = mass * specific heat * change in temperature

q = (65 grams water)(4.180 J/gC)(40o C - 25o C)

= 4075.5 joules (1 calorie/4.184 joules)

= 974 calories

==========

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12y ago

q = C x m x (Tf-Ti), where q is heat, C is specific heat, and (Tf-Ti) is change in temperature.

CH2O(l) = 4.184 J/goC

(Tf-Ti) = 45oC-30oC = 15oC

massH2O = 100.0g

q = 4.184 J/goC x 100.0g x 15oC = 6300J (rounded to two significant figures)

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11y ago

Use,

q(joules) = mass * specific heat * change in temperature

q = (10 grams water)(4.180 J/g C)(40o C - 20o C)

= 836 joules

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6y ago

The necessary heat is 15 calories (62,802 Joules).

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12y ago

ewfewf fw

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Q: How much heat does 32.0 g of water absorb when it is heated from 25.0 Celsius to 80.0 Celsius in calories?
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