# Study Guide 18 - Heat

- Calculate the intensity of the radiated energy from a body at 100ºC. Then calculate the net intensity if the surroundings are at 90ºC.

The total radiated intensity of a black body is given most simply by Stefan's Law

\(\mathrm {T = \sigma T^4,}\) where

\(\mathrm {Stefan's\; constant = \sigma = 5.672\times 10^{-8} Wm^{-2}T^{-4} \\ I = 5.672 \times 10^{-8}(100 + 273) = 1098\; W/m^2}\)

\(\mathrm {If \;the\; surroundings\; are\; at \;a \;temperature\; of\; 90^\circ C = 90 + 273 = 363\; K \\ I = \sigma(T_1{^4} - T_2{^4}) = 5.672 \times 10-8(373^4 - 363^4) = 113\; W/m^2}\)

- Two aluminum spheres are at a temperature of \(\mathrm {100^\circ C}\) in surroundings at \(\mathrm {90^\circ C.}\) The smaller sphere has a surface area of \(\mathrm{1\; m^2.}\) The larger sphere has 8 times the mass of the smaller. What is the ratio of the initial rate of heat loss of the large sphere to that of the small one?

Since the temperature of both spheres is the same, their ratio of heat loss is initially independent of T and depends only on their surface areas.

\(\mathrm {\frac{I _1}{I_2} = \frac{A_1}{A_2} \\ but\\ \frac{A_1}{A_2} = \bigg(\frac{R_2}{R_2}\bigg)^2 = \bigg(\frac{R_1^3}{R_2^3}\bigg)^{2/3}= \bigg(\frac{V_1}{V_2}\bigg)^{2/3}= \bigg(\frac{M_2}{M_2}\bigg)^{2/3}\\ Therefore\\ \frac{I_1}{I_2} = \bigg(\frac{M_1}{M_2}\bigg)^{2/3} = 8^{2/3} = 4}\)

- Two different hot objects, A and B, both at 100ºC, are placed in a room at 20ºC. After 20.0 minutes, A has cooled down to 70.0ºC and B to 80.0ºC

(a) Which one has the greater decay constant?

The one that cools faster has the greater decay constant, i.e., A.

(b) What is the ratio of the larger decay constant to the smaller?

For A:

\(\mathrm {T_A - T_S = (T_{0A} - T_{0S}) e^{-k_At} \\ 70 - 20 = (100 -20) e^{-20 k_A} \\ 50 = 80\; e^{20k_A} \\ - 20 k_A = \ell \;5/8 \\ k_A = 0.0235 \; min^{-1}}\)For B:

\(\mathrm {80 - 20 = (100 - 20) e^{-20k_B} \\ k_B = 0.0144 \;min ^{-1} \\ \frac{k_A}{k_B} = \frac{0.0235}{0.0114} = 1.63}\)

(c) How long will it take A to cool down from 70.0º to 40.0º?

\(\mathrm {40 - 20 = (70 - 20) e^{-0.0235 \ell}\\ 20 = 50 e^{- 0.0235t} \\ \ell n 2/5 = - 0.0235t \\ t = 3.9 \; min}\)

(d) How long to cool down from 40.0º to 20.0º?

It will take an infinitely long time to cool to the same temperature as the surroundings. (In practise-a very long but finite time.)