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3 years ago

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You pour 250 g of tea into a Styrofoam cup, initially at 80∘C and stir in a little sugar using a 100-g aluminum 20∘C spoon and l

eave the spoon in the cup. Assume the specific heat of tea is 4180 J/kg⋅∘C and the specific heat of aluminum is 900 J/kg⋅∘C. Part APart complete What is the highest possible temperature of the spoon when you finally take it out of the cup?

1 answer:

jok3333 [9.3K]3 years ago

3 0

To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.

By definition we know that the heat lost must be equal to the heat gained, ie

$Q_g = Q_l$

Where,

Q = Heat exchange

The heat exchange is defined as

$Q = c_p m \Delta T$

Where,

$c_p =$ Specific heat

m = mass

$\Delta T=$ Change in Temperature

Therefore replacing we have that

$Q_g = Q_l$

$c_{p-tea} m \Delta T = c_{p-al} m \Delta T$

Replacing with our values we have that

$0.25*4180*(80-T) = 0.1*900*(T-20)$

$11.61*(80-T) = T-20$

$T= \frac{948.8}{11.61}$

$T = 75.24\°C$

Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C

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2 years ago

Which term describes a gap in the geologic record that occurs when sedimentary rocks cover an eroded surface?

Lemur [1.5K]

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3 years ago

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A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

8 0

3 years ago

Select the correct answer.

Natali [406]

its b hoped i helped

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3 years ago

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Student swings a small rubber stopper attached to a string over her head in a horizontal, circular path. The string is 1.50 mete

harkovskaia [24]

Answer:

v = 18.84 m/s

Explanation:

Given that,

The length of the string, r = 1.5 m (it will act as radius)

The rubber stopper makes 120 complete circles every minute.

Since, 1 minute = 60 seconds

It means, its frequency is 2 circles every second.

Let we need to find the average speed of the rubber stopper. It can be calculated as follows :

$v=\dfrac{d}{T}$

d is distance, $d=2\pi r$ and 1/T = f (frequency)

$v=2\pi rf\\\\=2\pi \times 1.5\times 2\\\\=18.84\ m/s$

So, the average speed of the rubber stopper is 18.84 m/s.

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2 years ago