A 110.-g sample of copper (specific heat capacity 0.20 J/C g) is heated to 82.4C and then placed in a container of water at 22.3C. The final temperature of the water and copper is 24.9C. What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water.

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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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

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