Question

A Styrofoam cup (k = 0.010 W/(m∙ o C)) has cross-sectional area (A) of 3.0 x 10 −2m 2 . The cup is 0.589 cm thick (L). The temperature (T2) of the coffee is 86.0 o C. The air temperature in the room (T1) is 24.0 o C. Calculate the rate of conductive heat transfer (Q) in Watts.

Answer 1

Answer:

The rate of conductive heat transfer Q/t is approximately 3.158 Watts

Explanation:

The given parameters of the Styrofoam cup are;

The thermal conductivity of the cup, k = 0.010 W/(m·°C)

The cross-sectional area of the cup, A = 3.0 × 10⁻² m²

The thickness of the cup, L = 0.589 cm = 0.00589 m

The temperature of the coffee (in the cup) = 89°C, T₂ = 86.0 °C

The temperature of the air in the room (of the cup of coffee), T₁ = 24.0°C

The rate of conductive heat transfer of the cup of coffee, Q/t, is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1)}{L}$

Plugging in the value of the variables, in the above equation, gives;

$\dfrac{Q}{t} = \dfrac{0.010 \, W/(m\cdot ^{\circ }C) \times 3.0 \times 10^{-2} m^2 \times (86.0 ^{\circ} C - 24.0^{\circ}C)}{0.00589 \, m} = \dfrac{60}{19} \ W$

The rate of conductive heat transfer of the cup of coffee, Q/t = 60/19 W = $3.\overline {157894736842105263}$ W ≈ 3.158 W.