Ask question

Ask question

All categories

3 years ago

12

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 tempe

rature (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.

1 answer:

saveliy_v [14]3 years ago

4 0

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.

You might be interested in

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

ale4655 [162]

Answer:

<em>a) 50 J/kg</em>

<em>b) 721 67 KW</em>

<em></em>

Explanation:

The velocity of the wind v = 10 m/s

diameter of the blades d = 70 m

efficiency of the turbine η = 30%

density of air ρ = 1.25 kg/m^3

The area of the blade A = $\pi d^2/4$

A = $\frac{3.142 * 70^2}{4}$ = 3848.95 m^2

The mechanical energy air per unit mass is gives as

e = $v^2/2$ = $\frac{10^2}{2}$ = <em>50 J/kg</em>

<em></em>

Theoretical Power of the turbine P = ρAve

where

ρ is the density of air

A is the area of the blade

v is the velocity of the wind

e is the energy per unit mass

substituting values, we have

P = 1.25 x 3848.95 x 10 x 50 = 2405593.75 W

Actual power = ηP

where η is the efficiency of the turbine

P is the theoretical power of the turbine

Actual power = 0.3 x 2405593.75 = 721678.1 W

==> <em>721 67 KW</em>

7 0

3 years ago

Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter ???

Andrej [43]

We need to define the variables,

So,

$F_x (x) = 1-e^{-\lambda x}\\F_x (x) = 1-e^{-0.5x}$

Therefore, the probability that the repair time is more than 4 horus can be calculate as,

$P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018$

The probability that the repair time is more than 4 hours is 0.136

b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,

$P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}$

$P(x\geq 12|x>7)=0.6321$

The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63

3 0

3 years ago

Suppose you have a 9.00 V battery, a 2.00 μF capacitor, and a 7.40 μF capacitor. (a) Find the charge and energy stored if the ca

Andru [333]

Answer:

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

Explanation:

<u>a)</u>

<u>Identify the unknown: </u>

The charge and energy stored if the capacitors are connected in series

<u>List the Knowns: </u>

Capacitance of the first capacitor: $C_{1}$ = 2цF = 2 x 10-6 F

Capacitance of the second capacitor $C_{2}$ = 7.4цF = 7.4 x 10-6 F

Voltage of battery: V = 9 V

<u>Set Up the Problem: </u>

Capacitance of a series combination:

$\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }$ +............

$\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\$

Capacitance of a series combination is given by:

$C_{s}=\frac{Q}{V}$

Then the charge stored in the series combination is:

$Q=C_{s} V$

Energy stored in the series combination is:

$U_{c}=\frac{1}{2} V^{2} C_{s}$

<u>Solve the Problem: </u>

$Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J$

<u>b)</u>

<u>Identify the unknown: </u>

The charge and energy stored if the capacitors are connected in parallel

<u>Set Up the Problem: </u>

Capacitance of a parallel combination:

$C_{p} =C_{1} +C_{2} +C_{3}$

$C_{p} =2+7.4=9.4*10^-6F$

Capacitance of a parallel combination is given by

$C_{p} =\frac{Q}{V}$

Then the charge stored in the parallel combination is

$Q=C_{p} V$

Energy stored in the parallel combination is:

$U_{c}=\frac{1}{2} V^2C_{p}$

<u>Solve the Problem: </u><em> </em>

$Q=9.4*10^-6*9=8.46*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *9.4*10^-6=3.81*10^-4J$

5 0

3 years ago

Read 2 more answers

A panel crimper cuts metal True or False

densk [106]

Answer:

True (They can cut through most metal panels, and probably heavier)

Explanation:

Hope this helps you as much as intended.

3 0

3 years ago

Technician A says that the starter solenoid switches the high current on and off. Technician B says that the solenoid on the sta

jek_recluse [69]

Technician A is incorrect so that means Technician B is correct.

3 0

2 years ago