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Specific Heat

q = m x C x (Tf - Ti)

q = amount of heat energy gained or lost by substance = 1072 J

m = mass of sample in grams = 4.91g

C = heat capacity (J/g•oC) = 0.240 J/g•oC

Tf = final temperature = ? (melting point of silver)

Ti = initial temperature = 25oC

To solve for Tf:

q/m•C = Tf - Ti

Ti + q/m•C = Tf

25oC + 1072 J/(4.91g x 0.250 J/g•oC) = Tf

9.0 x 102 oC = Tf

(898oC rounded to two significant figures due to 25oC.)

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9y ago
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11y ago

q=mcdeltaT

q=(4.37)(0.24g)(25.0 degrees C-27.5 degrees C)

= -2.62 J

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12y ago

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

q = (4.37 g)(0.237 J/gC)(27.5o C - 25.0o C)

= 2.59 Joules

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7y ago

The necessary heat is: Q = 400 x 45 x 0,233 = 4,194 kJ

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Q: The specific heat of silver is 0.24 mc015-1jpg How many joules of energy are needed to warm 4.37 g of silver from 25.0mc015-2jpgC to 27.5mc015-3jpgC?
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