Heat lost by tubing=Heat gained by water+vessel
446×0.387×(89.5-T)=159×1×(T-22.5)+10×(T-22.5)
15447.8-172.6T=169T-3802.5
T=56.35°C(approx).
6Q11 A 446-g piece of copper tubing is heated to 89.5 degree C and placed in...
Q11 A 446-g piece of copper tubing is heated to 89.5degree C and placed in an insulated vessel containing 159 g of water at 22.8degree C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/degree C, what is the final temperature of the system (c of copper = 0.387 J/g middot degree C)? degree C
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I am currently working 6.45. I had the equation set up corrctly until i got to the circled portion of the second picture. I dont understand why they multiplied the (10J/C) by (Tfinal - 22.8C) when i already had (Tfinal -22.8C) in the equation right before that. I hope that makes sense. Thank you! place 6.45 A 455-g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming...
A 45.90 g sample of pure copper is heated in a test tube to 99.40°C. The copper sample is then transferred to a calorimeter containing 61.04 g of deionized water. The water temperature in the calorimeter rises from 24.47°C to 29.10°C. The specific heat capacity of copper metal and water are J J 0.387 and 4.184 respectively. gr°C g. °C Assuming that heat was transferred from the copper to the water and the calorimeter, determine the heat capacity of the...