Answer:
The answer to the question is
The mass of the water in the container, assuming that all the heat lost by the copper is gained by the water is 116.23 grams
Explanation:
To solve this we list out the variables thus,
mass of copper = 110g
specific heat capacity of copper = 0.20 j/C g
Initial temperature of copper = 82.4 C
Final temperature of copper and water = 24.9 C
therefore
From the formula for sensible heat, ΔH = m×c×ΔT
we have, where
H = senced heat,
m = mass,
ΔT = temperature change
110 × 0.20 × (82.4 - 24.9) = mw × 4.186 × (24.9 - 22.3)
1265 J = mw ×10.8836 or mw = 116.23 g
Therefpre the mass of water is 116.23 g
The mass of the water in the container assuming that all the heat lost by the copper is gained by the water is 116.29 g
We'll begin by calculating the heat lost by the copper. This can be obtained as follow:
Mass of copper (M) = 110 g
Specific heat capacity of copp(C) = 0.20 J/gºC
Temperature of copper (T꜀) = 82.4 °C
Equilibrium temperature (Tₑ) = 24.9 °C
Q = MC(T꜀ – Tₑ)
Q = 110 × 0.2 × (82.4 – 24.9)
Finally, we shall determine the mass of the water in the container. This can be obtained as follow:
Heat lost (Q) = Heat gained = 1265 J
Specific heat capacity of water (C) = 4.184 J/gºC
Equilibrium temperature (Tₑ) = 24.9 °C
Temperature (Tᵥᵥ) = 22.3 °C
Q = MC(Tₑ – Tᵥᵥ )
1265 = M × 4.184 × (24.9 – 22.3)
1265 = M × 4.184 × 2.6
1265 = M × 10.8784
Divide both side by 10.8784
M = 1265 / 10.8784
Thus, the mass of water in the container is 116.29 g
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