A cubical piece of heat-shield-tile from the space shuttle measures 0.11 m on a
ID: 1590556 • Letter: A
Question
A cubical piece of heat-shield-tile from the space shuttle measures 0.11 m on a side has thermal conductivity of 0.065 J/(s.m.C^degree). The outer surface of the tile is heated to a temperature of 1210^degree C, while the inner surface is maintained at a temperature of 17^degree C. How much heat flows from the outer to the inner surface of the tile in 4.0 minutes? If this amount of heat were transferred to 2.4 liters (2.4 kg) of liquid water, by how many Celsius degrees would the temperature of the water rise?Explanation / Answer
(a) Heat conduction Q/ Time = (Thermal conductivity) x (Area) x (Thot - Tcold)/Thickness
Heat conduction Q = (Thermal conductivity) x (Area of cubical piece side) x (Thot - Tcold) x (Time)/Thickness(the thickness of cubical piece = 0.11m)
k = 0.065 J/(s·m·C°).
Q = k* A*T* t / L
Q = 0.065 J/(s m · C°) x (0.11 m x 0.11 m) x (1210° C - 17° C) x (240 second) / (0.11m)
Q = 2047.188 J
(b) Q = (mass of water) x (specific heat of water) x (rise of temperature)
Q = m * C * T
T = Q / C m = 2047.188 J / (4186 J/(kg·C°) * 2.4 kg) = 0.203C°