## Heat added to water = (Heat absorbed by liquid) + (Heat

Heat added to water = (Heat absorbed by liquid) +
(Heat transferred to liquid from water (3)) +
+ (Heat loss from water container to air (2))
+ (Heat loss from liquid container to air (1))
+ (Heat loss from tube for 280mm)
+ (Heat loss from dish)

i have to find out, to what temperature water should be heated to maintain liquid at 312K.

whether applying this equation and substitution like below helps to find out the required water temperature.
ambient temperature 300K.

0.3× 4186 X (θ -300) = 0.0012 х1025 × 3600× (312 – 300) +
[60.8 (θ -312)+ 90.72 +28.8 (θ -300) + 40.32 + 7.9]~~~~~these are losses considering the inner temperature is 312K

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 Thread Tools

 Similar Threads for: Heat added to water = (Heat absorbed by liquid) + (Heat Thread Forum Replies Advanced Physics Homework 0 Science Textbook Discussion 9 Science Textbook Discussion 1 Introductory Physics Homework 7 Academic Guidance 10