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ycow [4]
2 years ago
13

Five-gram samples of copper and aluminum are at room temperature. Both receive equal amounts of energy due to heat flow. The spe

cific heat capacity of copper is 0.09 cal/g°C, and the specific heat capacity of aluminum is 0.22 cal/g°C. Which of the following statements is true?
a. The temperature of each sample increases by the same amount.
b. The aluminum will get hotter than the copper.
c. The copper will get hotter than the aluminum.
d. The temperature of each sample decreases by the same amount
Physics
1 answer:
hammer [34]2 years ago
5 0

Answer:

  • <u><em>c. The copper will get hotter than the aluminum.</em></u>

Explanation:

The <em>energy due to heat flow</em>, i.e. the heat energy, is proportional to the product of the mass, the <em>specific heat capacity</em>, and the change in <em>temperature</em>:

          Heat=mass\times \text{specific heat capacity }\times\Delta T

From which you obtain:

     \Delta T=\dfrac{Heat}{mass\times \text{specific heat capacity}}

That equation tells that the change in temperature is inversely related to the product of the mass and the specific heat.

In the problem, the masses of the samples of copper and aluminum are equal (<em>5.00g</em>) and both samples receive <em>equal amounts of energy due to heat flow</em>, thus the only difference is the <em>specific heat capacity</em> of each sample.

From the above stated relationship between the change in temperature, the heat, the mass, and the specific heat capacity, under the assumption of all the other conditions equal (heat energy and mass), the higher the specific heat capacity the lower the change in temperature, and the lower the specific heat capacity the greater the change in temperature.

The heat capacity of copper (0.09 cal/g°C) is lower than the specific heat capacity of aluminum is (0.22 cal/g°C), thus the increase in temeperature of the copper sample will be greater than that of the aluminum sample. This means that <em>the copper will get hotter than the aluminum (option c.)</em>

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Which of the following statement is not true about machines?
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3 0
2 years ago
A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate
yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

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Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

W=Pm\DeltaV

W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}

W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})

Where,R = gas constant

T = temperature

P = pressure

P_{atm}=Atmosphere pressure

m = mass

Put the value into the formula

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6 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
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