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zvonat [6]
3 years ago
12

What is the change of entropy for a heat engine using 500.0j at 20.0c

Mathematics
2 answers:
ExtremeBDS [4]3 years ago
8 0

The formula to find change in entropy is given by :

Change in entropy = \frac{Q}{T}

where Q is change in heat and T is temperature.

We are given Q = 500 joule and T = 20 degree C

Plugging these values in the formula,

Change in entropy = \frac{500}{20}

We get change in entropy = 25 J/Celsius.


Natasha_Volkova [10]3 years ago
3 0
1.71 J/K 
hope it helps :D!!
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Since the other angle that right next to it, the same angle you equal both angles together.

4x = 68.

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4x = 68
/4      /4

4 divided by 4 is 1 and 68 divided by 4 is 17

x = 17 is the answer.
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Answer:

A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.

B.)<u><em>The advantage of attempting to remove the block effect is</em></u>

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

Step-by-step explanation:

<em><u /></em>

<em><u>Using Two-way ANOVA method</u></em>

Given problem

<em><u>Observation              I          II       III          Row total (xr)</u></em>

A                              18 21 20            59

B                            24 26 27             77

C                            30 29 34             93

D                            22 25 24            71

<u>E                            20 23 24           63                      </u>

Col total (xc)             114 124 129        367

∑x²=9233→(A)

∑x²c/r

=1/5(114²+124²+129²)

=1/5(12996+15376+16641)

=1/5(45013)

=9002.6→(B)

∑x²r/c

=1/3(59²+77²+93²+71²+67²)

=1/3(3481+5929+8649+5041+4489)

=1/3(27589)

=9196.3333→(C)

(∑x)²/n

=(367)²/15

=134689/15

=8979.2667→(D)

Sum of squares total

SST=∑x²-(∑x)²/n

=(A)-(D)

=9233-8979.2667

=253.7333

Sum of squares between rows

SSR=∑x²r/c-(∑x)²/n

=(C)-(D)

=9196.3333-8979.2667

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Sum of squares between columns

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=(B)-(D)

=9002.6-8979.2667

=23.3333

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Source                 Sums         Degrees      Mean Squares

of Variation       of Squares   of freedom

<u>                               SS                 DF              MS       F p-value</u>

B/ w     SSR=217.0667              4 MSR=54.2667    32.56 0.0001

rows

B/w     SSC=23.3333         c-1=2 MSC=11.6667        7 0.01

columns

<u>Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667                  </u>

<u>Total SST=253.7333 rc-1=14                                                        </u>

Conclusion:

<u> 1. F for between Rows</u>

The critical region for F(4,8) at 0.05 level of significance=3.8379

The calculated F for Rows=32.56>3.8379

Therefore H0 is rejected

<u>2. F for between Columns</u>

The critical region for F(2,8) at 0.05 level of significance=4.459

We see that the calculated F for Colums=7>4.459

therefore H0 is rejected,and concluded that there is significant differentiating between columns

<u><em>Part B:</em></u>

To analyze the data for completely  randomized designs click on anova two factor without replication  in the data analysis dialog box of the excel spreadsheet.

The following table is obtained

Source DF             Sum                  Mean           F Statistic

<u>                 (df1,df2)    of Square (SS) Square (MS)                    P-value</u>

Factor A       1 1496.5444 1496.5444 769.6514          0.001297

Rows

Factor B -     2 19.4444           9.7222               5                  0.1667

Columns

Interaction

AB               2    3.8889   1.9444        0.1013         0.9045

<u> Error     12   230.4            19.2                                           </u>

<u>Total 17 1750.2778 102.9575                                                         </u>

<u />

<u>Factor - A- Rows</u>

Since p-value < α, H0 is rejected.

<u>Factor - B- Columns</u>

Since p-value > α, H0 can not be rejected.

The averages of all groups assume to be equal.

<u>Interaction AB</u>

Since p-value > α, H0 can not be rejected.

<u><em>The advantage of attempting to remove the block effect is</em></u>

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

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