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erma4kov [3.2K]
3 years ago
13

Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

ocks A and B are placed in contact, isolated, and allowed to come into equilibrium. Then they are placed in contact with block C. At that instant:a.TA = TB < TCb.TA = TB = TCc.TA = TB > TCd.TA + TB = TCe.TA - TB = TC
Physics
1 answer:
shepuryov [24]3 years ago
8 0

Answer:

b) TA = TB = TC

Explanation:

  • When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
  • During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
  • The equilibrium condition will be reached when the following equation be met:

       \Delta Q = c_{st}* m_{A} * (T_{fin}  - T_{0A} ) = c_{st}* m_{B} * (T_{0B}  - T_{fin} )

  • Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for  Tfin, as follows:

       (400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\  2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC

  • So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
  • When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
  • So, option b) is the right one.
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(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

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(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

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(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

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