Answer :
- The final temperature of the system after the copper tubing is heated and placed in water is approximately 26.6°C.
1. Given:
- Mass of copper tubing ([tex]m_{copper[/tex]) = 505 g
- Initial temperature of copper ([tex]T_{copper[/tex]) = 99.9°C
- Mass of water ([tex]m_{water[/tex]) = 59.8 g
- Initial temperature of water ([tex]T_{water[/tex]) = 24.8°C
- Heat capacity of copper ([tex]c_{copper[/tex]) = 0.387 J/g·K
- Heat capacity of vessel ([tex]c_{vessel[/tex]) = 10.0 J/K
2. Calculate the heat lost by the copper tubing and gained by the water:
- Heat lost = Heat gained
-[tex]m_{copper} * c_{copper} * (T_{final} - T_{copper}) = m_{water} * c_{water} * (T_{final} - T_{water})[/tex]
3. Solve for the final temperature (T_final):
505g * 0.387 J/g·K * ([tex]T_{final[/tex] - 99.9°C) = 59.8g * 4.18 J/g·K * ([tex]T_{final[/tex] - 24.8°C)
- 195.435 ([tex]T_{final[/tex]- 99.9) = 250.364 ([tex]T_{final[/tex]- 24.8)
- 195.435 [tex]T_{final[/tex]- 19500.865 = 250.364[tex]T_{final[/tex]- 6209.312
- 54.929[tex]T_{final[/tex] = 13291.553
- [tex]T_{final[/tex]≈ 26.6°C
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Question
A 505-gram piece of copper tubing is heated to 99.90 degrees Celsius and placed in an insulated vessel containing 59.8 grams of water at 24.8 degrees Celsius. Assuming no loss of water and that the vessel has a heat capacity of 10.0 J/K, what is the final temperature of the system? (specific heat capacity of copper = 0.387J/g degree C)
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