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PIT_PIT [208]
3 years ago
9

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

heat? a. 50.0 g Fe, CFe = 0.449 J/g°C b. 50.0 g water, Cwater = 4.18 J/g°C c. 25.0 g Pb, CPb = 0.128 J/g°C d. 25.0 g Ag, CAg=0.235 J/g°C e. 25.0 g granite, Cgranite = 0.79 J/g°C
Chemistry
1 answer:
Brut [27]3 years ago
5 0

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

q=mC\times \Delta T      ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

\Delta T = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

  • <u>Option a:</u>  50.0 g Fe, C_{Fe}=0.449J/g^oC

We are given:

m=50.0g\\C_{Fe}=0.449J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC

Change in temperature = 8.99°C

  • <u>Option b:</u>  50.0 g water, C_{water}=4.18J/g^oC

We are given:

m=50.0g\\C_{water}=4.18J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC

Change in temperature = 0.96°C

  • <u>Option b:</u>  25.0 g Pb, C_{Pb}=0.128J/g^oC

We are given:

m=50.0g\\C_{Pb}=0.128J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC

Change in temperature = 62.5°C

  • <u>Option d:</u>  25.0 g Ag, C_{Ag}=0.235J/g^oC

We are given:

m=25.0g\\C_{Ag}=0.235J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC

Change in temperature = 34.04°C

  • <u>Option e:</u>  25.0 g granite, C_{granite}=0.79J/g^oC

We are given:

m=25.0g\\C_{Fe}=0.79J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

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A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
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3.6667

Explanation:

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{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

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P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

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<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

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P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

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<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

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