How much will be the temperature of a cup (180 gram) of coffee at 95 degree centigrade be reduced when a 45 gram silver spoon (specific heat 0.25 / g degree celsius) at 25 degree celsius is placed in the coffee and the two are allowed to reach the same temperature assume that the coffee has the same density and specific heat as water.

The temperature of the cup of of coffee is reduced by 1 °C

Explanation:

The cup of coffee (system 1) and silver spoon (system 2) can be considered an isolated system for this problem, so the exchange of heat to the outside is zero:

Q₁ + Q₂ = 0 = m₁c₁ΔT₁ + m₂c₂ΔT₂

m₁c₁(Tfinal - Tinitial₁) + m₂c₂(Tfinal - Tinitial₂) = 0

The equation is rearranged to solve for Tfinal, which will be the same for both systems:

m₁c₁(Tfinal) - m₁c₁(Tinitial₁) + m₂c₂(Tfinal) - m₂c₂(Tinitial₂) = 0

m₁c₁(Tfinal) + m₂c₂(Tfinal) = m₁c₁(Tinitial₁) + m₂c₂(Tinitial₂)

Substituting in some values at this point gives:

(180g)(4.190Jg⁻¹K⁻¹)(Tfinal) + (45g)(0.25Jg⁻¹K⁻¹)(Tfinal) = (180g)(4.190Jg⁻¹K⁻¹)(95 + 273.15 K) + (45g)(0.25Jg⁻¹K⁻¹)(25 + 273.15 K)

(754.2JK⁻¹)(Tfinal) + (11.25JK⁻¹)(Tfinal) = 277658.73J + 3354.1875J

(765.45JK⁻¹)(Tfinal) = 281012.9175J

Tfinal = 367.12K = (367.12 - 273.15)C = 94C

The temperature of the cup of coffee was reduced by 1 degree Celsius

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