A 1.3 g sample of gold (specific heat capacity = 0.130 J/g °C) is heated using 86.9 J of energy. If the original temperature of the gold is 25.0°C, what is its final temperature?

556.95°C

Explanation:

The following were obtained from the question:

M = 1.3g

C = 0.130 J/g °C

Q = 86.9J

T1 = 25.0°C

T2 =?

Q = MC(T2 — T1)

86.9 = 1.3 x 0.13(T2 — 25)

Clear the bracket

86.9 = 0.169T2 — 4.225

Collect like terms

89.9 + 4.225 = 0.169T2

94.125 = 0.169T2

Divide both side by 0.169

T2 = 94.125/0.169

T2 = 556.95°C

-be a crystalline solid

- be soluble in water

-conduct electricity

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