A cubical piece of heat-shield-tile from the space shuttle measures 0.16 m on a side and has a thermal conductivity of 0.065 J/(s·m·C°). The outer surface of the tile is heated to a temperature of 1020°C, while the inner surface is maintained at a temperature of 19°C. (a) How much heat flows from the outer to the inner surface of the tile in 5.0 minutes? (b) If this amount of heat were transferred to 3.9 liters (3.9 kg) of liquid water, by how many Celsius degrees would the temperature of the water rise?
I have the answer to part a) 3123.12 J
B) is giving me issues, however
I have the answer to part a) 3123.12 J
B) is giving me issues, however
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a) heat flowing / time = KAdT/l
(K= themal conductivity , A= area , l = length)
So heat (Q) = KAdTxt/l= 0.065 x 0.16*2 x (1020-19) x 5 x 60 x 60 / 0.16 = 187387.2 J
B) So now this whole heat flowing out is supplied to 3.9 kg of water. So definitely it will rise temperature of it (it can further do more)
So, Q= ms (dT) ---(dT= change in temperature)
So 187387.2 = 3.9kg x 4200 J/kg-C x dT
Solving we get,, dT = 11.44 C
(K= themal conductivity , A= area , l = length)
So heat (Q) = KAdTxt/l= 0.065 x 0.16*2 x (1020-19) x 5 x 60 x 60 / 0.16 = 187387.2 J
B) So now this whole heat flowing out is supplied to 3.9 kg of water. So definitely it will rise temperature of it (it can further do more)
So, Q= ms (dT) ---(dT= change in temperature)
So 187387.2 = 3.9kg x 4200 J/kg-C x dT
Solving we get,, dT = 11.44 C