Question

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?

Answer 1

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

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:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

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

$-16.902\,\frac{J}{^{\textdegree}C}\cdot (T-135^{\textdegree}C) = 1063.498\,\frac{J}{^{\textdegree}C} \cdot (T-39^{\textdegree}C)$

$43758,192\,J = 1080.4\,\frac{J}{^{\textdegree}C}\cdot T$

The final temperature of the system is:

$T = 40.501\,^{\textdegree}C$

Answer 2

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