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26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated cup.What will
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Answers ( )
Answer:
The new water temperature is 26.4 °C
Explanation:
Given;
mass of copper,
= 26 g = 0.026 kg
temperature of copper, t = 300 °C
volume of water, V = 120 mL = 0.12 L
temperature of water, t = 21 °C
density of water, ρ = 1 kg/L
mass of water = density x volume
mass of water = (1 kg/L) x 0.12 L = 0.12 kg
heat lost by copper = heat gained by water
Both copper and water reach final temperature, T
Heat gained by water,
=
cΔθ = 
Heat lost by copper is given by;
504(T- 21) = 10.01(300 – T)
504 T – 10584 = 3003 – 10.01 T
504 T + 10.01 T= 3003 + 10584
514.01 T = 13587
T = (13587) / 514.01
T = 26.4 °C
Therefore, the new water temperature is 26.4 °C