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A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

- The specific heat of copper is 385 J/kg°C.
- The specific heat capacity of water is 4182 J/kg°C.
- The latent heat of fusion of ice is 33600 J/kg.
- The specific heat capacity of lead is 128 J/kg°C.

A) 34.6°C
B) 42.9°C
C) 51.3°C
D) 58.7°C

Answer :

In this case, the final temperature is approximately 1.809°C. Since this value is less than 0°C, it means that the ice will not completely melt. Instead, the final temperature will be 0°C, the temperature at which ice starts melting. Therefore, the final temperature of the system, including the lead, will be 0°C.

So, none of the options is correct.

To solve this problem, we need to consider the heat gained or lost by each component in the system. We'll use the principle of conservation of energy, where the heat lost by the lead equals the heat gained by the copper, water, and ice.

First, let's find the heat absorbed by the lead:

Q[tex]_{\text{lead}}[/tex] = m[tex]_{\text{lead}}[/tex] × c[tex]_{\text{lead}}[/tex] × (T[tex]_f[/tex] - T[tex]_{\text{initial}}[/tex])

Where:

  • m[tex]_{\text{lead}}[/tex]= 0.750 kg (mass of lead)
  • c[tex]_{\text{lead}}[/tex]= 128 J/kg°C (specific heat capacity of lead)
  • T[tex]_{\text{initial}}[/tex] = 255°C (initial temperature of lead)
  • T[tex]_f[/tex] (final temperature of the system, including the lead)

Q[tex]_{\text{lead}}[/tex] = 0.750 kg × 128 J/kg°C × (T[tex]_f[/tex] - 255°C)

Now, let's find the heat gained by the copper, water, and ice:

Heat gained by the copper:

Q[tex]_{\text{copper}}[/tex] = m[tex]_{\text{copper}}[/tex] × c[tex]_{\text{copper}}[/tex] × (T[tex]_f[/tex]- T[tex]_{\text{initial}}[/tex])

Where:

  • m[tex]_{\text{copper}}[/tex] = 0.100 kg (mass of copper)
  • c[tex]_{\text{copper}}[/tex] = 385 J/kg°C (specific heat capacity of copper)

Heat gained by the water:

Q[tex]_{\text{water}}[/tex] = m[tex]_{\text{water}}[/tex] × c[tex]_{\text{water}}[/tex] × (T[tex]_f[/tex]- T[tex]_{\text{initial}}[/tex])

Where:

  • m[tex]_{\text{water}}[/tex] = 0.160 kg (mass of water)
  • c[tex]_{\text{water}}[/tex] = 4182b J/kg°C (specific heat capacity of water)

Heat gained by the ice (to raise its temperature to 0°C:

Q[tex]_{\text{ice}}[/tex] = m[tex]_{\text{ice}}[/tex] × c[tex]_{\text{ice}}[/tex] × (0°C - T[tex]_{\text{initial}}[/tex])

Where:

  • m[tex]_{\text{ice}}[/tex] = 0.0180 kg (mass of ice)
  • c[tex]_{\text{ice}}[/tex] = 4182 J/kg°C (specific heat capacity of ice)

Heat gained by the ice to melt it completely into water:

Q[tex]_{\text{fusion}}[/tex] = m[tex]_{\text{ice}}[/tex] × Lf

Where:

Lf = 33600 J/kg (latent heat of fusion of ice)

Since there is no change in the temperature of the water and ice once they reach 0°C, the heat gained by them can be calculated with the latent heat of fusion directly.

Now, according to the principle of conservation of energy:

[tex]\[ Q_{\text{lead}} = Q_{\text{copper}} + Q_{\text{water}} + Q_{\text{ice}} + Q_{\text{fusion}} \][/tex]

Substituting the expressions we derived earlier, we get:

0.750 × 128 × (T[tex]_f[/tex] - 255) = 0.100 × 385 × (T[tex]_f[/tex] - 255) + 0.160 × 4182 × (T[tex]_f[/tex] - 255) + 0.0180 × 4182 × (T[tex]_f[/tex] - 255) + 0.0180 × 33600

Now, we solve for T[tex]_f[/tex]. This will give us the final temperature of the system.

Let's simplify the equation and solve for T[tex]_f[/tex]:

96(T[tex]_f[/tex] - 255) = 38.5(T[tex]_f[/tex]- 255) + 669.12(T[tex]_f[/tex] - 255) + 75.276(T[tex]_f[/tex] - 255) + 604.8

96T[tex]_f[/tex] - 24480 = 38.5T[tex]_f[/tex] - 9877.5 + 669.12T[tex]_f[/tex]- 17037.43 + 75.276T[tex]_f[/tex] - 1920.78 + 604.8

Now, let's combine like terms:

96T[tex]_f[/tex]- 38.5T[tex]_f[/tex] - 669.12T[tex]_f[/tex] - 75.276T[tex]_f[/tex] = 24480 - 9877.5 - 17037.43 - 1920.78 + 604.8

309.324T[tex]_f[/tex] = 559.09

Now, solve for T[tex]_f[/tex]:

T[tex]_f[/tex] = 559.09 / 309.324

T[tex]_f[/tex] ≈ 1.809

So, the final temperature is approximately 1.809°C.

Therefore, the final temperature of the system, including the lead, will be 0°C.

So, none of the options is correct.

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