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

hamshie.k

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