Answer:
Option C. 30°C.
Explanation:
The following data were obtained from the question:
Mass of water (Mw) = 0.5 Kg
Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C
Initial temperature of water (Tw1) =
22°C
Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C
Mass of copper (Mc) = 0.5 Kg
Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C
Initial temperature of copper (Tc1) = 115°C
Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C
Final temperature (T2) =..?
Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.
Thus, we can determine the final temperature of the water and copper as follow:
Q = MwCwΔT + McCcΔT
0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)
0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)
0 = 2.09T2 – 45.98 + 0.193T2 – 22.195
Collect like terms
2.09T2 + 0.193T2 = 45.98 + 22.195
2.283T2 = 68.175
Divide both side by the coefficient of T2 i.e 2.283
T2 = 68.175/2.283
T2 = 29.8 ≈ 30°C
Therefore, the final temperature of water and copper is 30°C.
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