A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost to the surroundings, what will the final temperature of the system be?

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

[tex]-Q_{out,Cu} = Q_{in,H_{2}O}[/tex]

After a quick substitution, the expanded expression is:

[tex]-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)[/tex]

[tex]-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)[/tex]

[tex]43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T[/tex]

The final temperature of the system is:

[tex]T = 40.501\,^{\textdegree}C[/tex]

asuwafohjames asuwafohjames · Answer 2 · 2020-03-17T05:07:02+01:00

Answer:

40.497 °C

Explanation:

Heat lost by copper = heat gained by water

CM(T₁-T₃) = cm(T₃-T₂)........................ Equation 1

Where C = specific heat capacity of copper, M = mass of copper, T₁ = Initial Temperature of copper, T₂ = Initial Temperature of water, T₃ = final temperature of the system,

Make T₃ the subject of the equation

T₃= (CMT₁+cmT₂)/(CM+cm)................. Equation 2

Given: C = 0.385 J/g°C, M = 43.9 g, c = 4.2 J/g°C, m = 254 g, T₁ = 135 °C, T₂ = 39 °C

Substitute into equation 2

T₃ = (0.385×43.9×135+4.2×254×39)/(0.385×43.9+4.2×254)

T₃ = (2281.70+41605.2)/(16.9015+1066.8)

T₃ = 43886.9/1083.7015

T₃ = 40.497 °C.

Hence the final temperature of the system = 40.497 °C