Deduce whether the metal block was made from copper or tin. The metal block has a mass of 220g and was placed in boiling water at 100°C. It was then transferred into 300g of water... Deduce whether the metal block was made from copper or tin. The metal block has a mass of 220g and was placed in boiling water at 100°C. It was then transferred into 300g of water at 19°C inside a glass container of mass 50g. The final temperature of the water was 23°C. Use the following values: specific heat capacity of water = 4200J kg⁻¹K⁻¹, specific heat capacity of glass = 840J kg⁻¹K⁻¹, specific heat capacity of copper = 390 J kg⁻¹K⁻¹, specific heat capacity of tin = 230 J kg⁻¹K⁻¹. Read more

Steps to Solve

1. Calculate the heat gained by the water

Use the formula $Q = mc\Delta T$, where $Q$ is the heat gained, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.

For the water: $m_{water} = 300g = 0.3kg$ $c_{water} = 4200 J kg^{-1} K^{-1}$ $\Delta T_{water} = 23°C - 19°C = 4°C$

$Q_{water} = (0.3 kg) \times (4200 J kg^{-1} K^{-1}) \times (4°C) = 5040 J$

2. Calculate the heat gained by the glass container

Use the same formula $Q = mc\Delta T$ for the glass container.

$m_{glass} = 50g = 0.05kg$ $c_{glass} = 840 J kg^{-1} K^{-1}$ $\Delta T_{glass} = 23°C - 19°C = 4°C$

$Q_{glass} = (0.05 kg) \times (840 J kg^{-1} K^{-1}) \times (4°C) = 168 J$

3. Calculate the total heat gained by the water and the glass container

$Q_{total} = Q_{water} + Q_{glass} = 5040 J + 168 J = 5208 J$

4. Calculate the heat lost by the metal block

The heat lost by the metal block is equal to the heat gained by the water and the glass container (conservation of energy).

$Q_{metal} = 5208 J$

5. Calculate the specific heat capacity of the metal block

Use the formula $Q = mc\Delta T$ to solve for $c$, where $Q$ is the heat lost, $m$ is the mass of the metal, and $\Delta T$ is the change in temperature of the metal.

$m_{metal} = 220g = 0.22kg$ $\Delta T_{metal} = 100°C - 23°C = 77°C$

$Q_{metal} = m_{metal} \times c_{metal} \times \Delta T_{metal}$ $5208 J = (0.22 kg) \times c_{metal} \times (77°C)$

$c_{metal} = \frac{5208 J}{(0.22 kg) \times (77°C)} = \frac{5208}{16.94} J kg^{-1} K^{-1} \approx 307.44 J kg^{-1} K^{-1}$

6. Compare the calculated specific heat capacity with the given values for copper and tin

The calculated specific heat capacity of the metal block is approximately $307.44 J kg^{-1} K^{-1}$. Comparing this with the given values:

• Copper: $390 J kg^{-1} K^{-1}$
• Tin: $230 J kg^{-1} K^{-1}$

The calculated value is closer to copper than tin.

7. Conclusion

The metal block is more likely to be copper.