# Question 97470

Dec 21, 2016

It is clear that the slab of steel at ${70}^{\circ} C$ will loose heat via all three modes: Conduction, Convection and Radiation simultaneously and finally will be at ambient temperature given as ${10}^{\circ} C$.

$\text{Total heat lost by the steel slab} = m s t$
where $m$ is mass of slab, $s$ specific heat of steel and $t$ is change in its temperature.

Assuming density $\rho$ of steel$= 7850 k g {m}^{-} 3$, $s = 511 J k {g}^{-} 1 {K}^{-} 1$
$m = \rho \times \text{volume}$
$m = 7850 \times \left(0.01 \times 1 \times 3\right) = 235.5 k g$

$\therefore \text{Total heat lost by the steel slab} = 235.5 \times 511 \times \left(70 - 10\right) = 7220430 J$
-.-.-.-.-.-.-.-.-.-.-.-.-.-.

A. Now, rate of heat loss due to conduction $\dot{Q}$ from slab to deck is calculated as below.

Lets assume that the $1.0 c m$ thick slab is lying flat on the deck.

Deck is at temperature of surroundings$= {10}^{\circ} C$

The heat loss from top to bottom of slab is found by taking
ΔT = 70^@C – 10^@C = 60^@C = 60 K, the temperature difference between the slab and deck.

Area $A$ through which heat is lost$= 1 \times 3 = 3 {m}^{2}$

Thermal conductivity $k \left(\text{steel}\right) = 50 W {m}^{-} 1 {K}^{-} 1$

We know that basic conduction problem equation is

dot Q=(kAΔT)/"thickness of slab"

Inserting above values we get
$\dot{Q} = \frac{50 \times 3 \times 60}{0.01}$
$\implies \dot{Q} = 900000 W$

B. Similarly basic convection problem equation is
basic equation for convection,

dotQ=hAΔT
where heat transfer coefficient $h = 10 W {m}^{-} 2 {K}^{-} 1$.

Area $A$ in contact with air$= 2 \times 0.01 \left(3 + 1\right) + 3 \times 1 = 3.08 {m}^{2}$

Inserting above values we get
$\dot{Q} = 10 \times 3.08 \times 60 = 1848 W$

C. For radiation loss we have Stefan-Boltzmann Law.

dotQ=εAσ(T_1^4– T_2^4)
where $\sigma$ is the Stefan Boltzmann constant = 5.670367 xx 10^-8 Wm^-2 K^-4, ${T}_{1} K$ is temperature of radiating body and ${T}_{2} K$ is temperature of surroundings.

We are given the emissivity for steel, ε= 0.85. The radiating surface area as calculated above, $A = 3.08 {m}^{2}$
Inserting given values we get
dotQ=0.85xx3.08xx5.670367 xx 10^-8(343^4– 283^4)=1102W#

D. Total heat loss rate $= 900000 + 1848 + 1102 = 902950 W$

Assuming rate of heat loss via each mode is constant throughout the time of loss.
Heat lost via conduction$= 7220430 \times \frac{900000}{902950} \approx 7196840 J$