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42 J

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7y ago
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13y ago

Use this formula.

q(in Joules) = mass * specific heat of water * change in temp.(final - initial )

q = nCT

q = (46.0 grams)(4.180 J/gC)(100 C - 0 C)

= 19228 Joules

correct for significant figures

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11y ago

Q = mCΔt, where Q is energy added or lost in Joules, m is mass in grams, C is specific heat, Δtis change in temperature in oC.

m = 2.0g

Δt = 5.0 oC

CH2O = 4.186 Joules/gram oC

Q = 2.0g x 4.180J/g•oC x 5.0 oC = 42 Joules (rounded to two significant figures)

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10y ago

For each Celsius degree of heating, you must add 4.184 joules per gram,

or 192.4 joules for the 46 gram sample.

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12y ago

1 calories (or 4,184 joule) for 1 g and for each Celsius degree.

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12y ago

Q (heat) = mass*Cp*Temp. Change (where Cp is specific heat)

Q = 5.0 grams * 4.18 J/goC * (30-20oC)

=5*4.18*10 = 209 J

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Q: How many joules of energy are necessary to heat a sample of water with a mass of 46.0 grams from 0.0 degrees celsius to 100 degrees celsius?
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