when 200.0g of chromium is placed in 200 g of water at 25 degrees Celsius in a coffee cup calorimeter, 6690 j of heat is absorbed by water, the specific heat chromium is 0.449 j/g-k
I was confused when the entholpy's unit is kj/kg, i believe you meant to say kj/kg*k. For my book, the entholpy, h, = 119.48 btu/lbm and converting it to kj/kg is 277.91048 kj/kg. However, if you convert entholpy, h, = 119.48 btu/lbm to kj/kg*k is 500.2389 kj/kg*k.
from that 1st law equation she used, the change in velocity part should have been multiplied by an m_dot. then dividing mdot from both sides the mdot goes away
The conversion factor for ft^2/sec^2 to BTU/lb used in problems in Michael Moran's Fundamentals of Engineering Thermodynamics is (1lbf/32.2 lb-ft/sec^2)(1 BTU/778 ft-lbf). For example 1,125,000 ft^2/sec^2 would convert as follows: (1,125,000 ft^2/sec^2)(1lbf/32.2lb-ft/sec^2)(1 BTU/778 ft-lbf) = 44.9 BTU/lb.
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when 200.0g of chromium is placed in 200 g of water at 25 degrees Celsius in a coffee cup calorimeter, 6690 j of heat is absorbed by water, the specific heat chromium is 0.449 j/g-k
I was confused when the entholpy's unit is kj/kg, i believe you meant to say kj/kg*k. For my book, the entholpy, h, = 119.48 btu/lbm and converting it to kj/kg is 277.91048 kj/kg. However, if you convert entholpy, h, = 119.48 btu/lbm to kj/kg*k is 500.2389 kj/kg*k.
i am so glad for this channel, its helping me alot for college
Helpful video thanks
Well done. Thank you for the content. Much appreciated.
in my "thermodynamics and engineering approach ninth edition" this is 5-28 by the way people
When you read off the table to get 437K did you assume the outlet pressure was the same as the inlet pressure?
Hi all, does someone know why the outlet speed is provided? is it not possible to calculate it somehow from the other inputs? thanks
why u didnt say to use interpolation to find T2 ??? make me confuse :D
Hi. What is the reason for cancelling the mass?
from that 1st law equation she used, the change in velocity part should have been multiplied by an m_dot. then dividing mdot from both sides the mdot goes away
what
😅😅 Thanks
So what would the conversion factor be for ft^2/sec^2 to BTU/lb?
The conversion factor for ft^2/sec^2 to BTU/lb used in problems in Michael Moran's Fundamentals of Engineering Thermodynamics is (1lbf/32.2 lb-ft/sec^2)(1 BTU/778 ft-lbf). For example 1,125,000 ft^2/sec^2 would convert as follows: (1,125,000 ft^2/sec^2)(1lbf/32.2lb-ft/sec^2)(1 BTU/778 ft-lbf) = 44.9 BTU/lb.