Physics Asked on August 25, 2021
The question is:
"0.38kg of a liquid at 12 degrees C is heated in a copper can weighing 0.9kg by an electrical heater of power 20W for 3 minutes. It reaches a temperature of 17C. Assuming no heat loss to the surroundings, calculate a value for the specific heat capacity of the liquid."
Shc of can is 400Jkg^-1K^-1
So far I have the following equation:
DeltaQcan + DeltaQliquid = 20360
Using Q=mcDeltaT,
DeltaQcan = mcanccanDeltaTcan = 0.9400DeltaTcan
DeltaQliquid = 0.38cliquid5
The answer says cliquid = 946.4Jkg^-1K^-1, which you can achieve through saying DeltaTcan = 5, hence:
0.94005 + 0.38cliquid5 = 20360, solving for 947.4.
However, I am unsure about the temperature change of the can. It doesn’t say that the liquid is in thermal equilibrium with the can after the three minutes, so why can we assume that the can has the same temperature change as the liquid? Why can the can not be hotter than the water after those three minutes?
The can is made of copper, which has a high thermal (as well as electrical) conductivity. (Both are related to the mobility of electrons in the bulk metal.) Heat energy flows very easily and quickly into or out of it, if there is a difference in temperature between the can and anything it is in contact with - such as the water. So you can assume that it is at the same temperature as the water at all times.
Answered by sammy gerbil on August 25, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP