Question

A cubical piece of heat-shield-tile from the space shuttle measures 0.15 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 1300°C, while the inner surface is maintained at a temperature of 22°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 3.0 liters (3.0 kg) of liquid water, by how many Celsius degrees would the temperature of the water rise?

Answer 1

Explanations & Calculations

• Heat transfer through the insulating brick under steady state condition is considered here (assume all the heat enters from the 1300 C⁰ face exits at 20C⁰ face).
• Rate of heat transfer is given by $\small R =\frac{Q}{t}=\frac{kA(\Delta \theta )}{\Delta d}$.
• Area (A) of a side of the brick = $\small (0.15m)^2 = 0.0225m^2$

• Then the amount of heat transferred in 5s

$\qquad\qquad \begin{aligned} \small Q &= \small R\times5min\\ &= \small \frac{0.065\times 0.0225m^2\times(1300-22)C^0}{0.15m}\times(5\times60s)\\ &= \small \bold{3738.15J} \end{aligned}$

• If the temperature change of water is $\small \Delta\theta_1$ then,

$\qquad\qquad \begin{aligned} \small Q&= \small mC \Delta \theta\\ \small 3738.15J &= \small 3kg\times 4200J/(kgC)\times \Delta \theta\\ \small \bold{\Delta \theta} &= \small \bold{0.267C^0}(<1C^0) \end{aligned}$