A 2.575 g piece of gold (specific heat = 0.129 J/g°C) at a temperature of 75°C is placed into cold water at 10°C. If the gold loses 10.0 J of energy, what is its final temperature in the water? Enter your answer to the

nearest whole number of °C.

The final temperature of the mixture is 44.9°C

Explanation:

Mass of the substance (gold) = 2.575g

Specific heat capacity of gold = 0.129J/g°C

Initial temperature (T1) = 75°C

Final temperature (T2) = ?

Energy lost = 10J

Heat energy(Q) = MC∇T

Q = heat energy (in this case lost)

M = mass of the substance

C = specific heat capacity of the substance

∇T = change in temperature of the substance = (T2 - T1)

Q = MC∇T

Q = MC(T2 - T1)

-10 = 2.575 × 0.129 × (T2 - 75) energy is -ve because it was energy lost.

-10 = 0.3321 × (T2 - 75)

-10 = 0.3321T2 - 24.9075

Collect like terms

0.3321T2 = 24.9075 - 10

0.3321T2 = 14.9075

T2 = 14.9075 / 0.3321

T2 = 44.88

T2 = 44.9°C

The final temperature of the mixture is 44.9°C

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