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.

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;

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

The rate of conductive heat transfer of the cup of coffee, Q/t = 60/19 W = W ≈ 3.158 W.

the equation above shows how temperature f, measured in degrees fahrenheit, relates to a temperature c, measured in degrees celsius. based on the equation, which of the following must be true?

a temperature increase of 1 degree fahrenheit is equivalent to a temperature increase of

5

9

degree celsius.

a temperature increase of 1 degree celsius is equivalent to a temperature increase of 1.8 degrees fahrenheit.

a temperature increase of

5

9

degree fahrenheit is equivalent to a temperature increase of 1 degree celsius.

a) i only

b) ii only

c) iii only

d) i and ii only

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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