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2 years ago

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You are given a sample of metal and asked to determine its specific heat. you weigh the sample and find that its weight is 28.4

n. you carefully add 1.25 x 104 j of heat energy to the sample and find that its temperature rises to 18.0 °c. what is the sample’s specific heat?

1 answer:

ZanzabumX [31]2 years ago

7 0

The specific heat of the given sample of metal is 14.32 J/kg.K.

<h3>What is the Specific heat of a sample?</h3>

The specific heat of a sample is the amount of heat needed or required to raise the temperature of that sample by 1K. It is given by the formula:

Q = mCΔT

where;

Q = Heat transferred
m = Mass of the substance
C = Specific heat
ΔT= Change in temperature.

Recall that:

weight of a substance = mass × gravity

mass = 28.4 N/ 9.8 m/s²
mass = 3 N/m/s² = 3 kg

From the equation:

Q = mCΔT

$\mathbf{C = \dfrac{1.25 \times 10^4 \ J}{3 \ Kg \times 291 \ K }}}$

C = 14.32 J/kg.K

Learn more about specific heat here:

brainly.com/question/21406849

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Sampling distribution involves the proportions of a data element in a given sample.

<em>The proportion of Good TV set is 0.67</em>
<em>The number of ways of selecting 5 from 6 TV sets is 6</em>
<em>The number of ways of selecting 4 from 6 TV sets is 15</em>

<em />

Given

$n = 6$

Sample Space = Good, Good, Defective, Defective, Good, Good

<u>(a) Proportion that are good</u>

From the sample space, we have:

$Good = 4$

So, the proportion (p) that are good are:

$p = \frac{Good}{n}$

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<u>(b) Ways to select 5 samples (without replacement)</u>

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 5$

So, we have:

$^6C_5 = \frac{6!}{(6 - 5)!5!}$

$^6C_5 = \frac{6!}{1!5!}$

$^6C_5 = \frac{6 \times 5!}{1 \times 5!}$

$^6C_5 = \frac{6}{1}$

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First, we calculate the number of ways to select 4.

This is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where

$r = 4$

So, we have:

$^6C_4 = \frac{6!}{(6 - 4)!4!}$

$^6C_4 = \frac{6!}{2!4!}$

$^6C_4 = \frac{6 \times 5 \times 4}{2 \times 1 \times 4!}$

$^6C_4 = \frac{30}{2}$

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So, the table is as follows:

$\left[\begin{array}{ccc}TV&Good&Proportion\\1,2,3,4&2&0.5&2,3,4,5&2&0.5&3,4,5,6&2&0.5\\4,5,6,1&3&0.75&5,6,1,2&4&1&6,1,2,3&3&0.75\\1,2,3,5&3&0.75&3,5,6,2&3&0.75&1,3,4,5&2&0.5\\1,3,4,6&2&0.5&1,4,5,2&3&0.75&2,4,6,1&3&0.75\\2,4,6,3&2&0.5&2,4,6,5&3&0.75&3,5,6,1&3&0.75\end{array}\right]$

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$p = \frac{Good}{n}$

<u>(d) The sampling distribution</u>

In (a), we have:

$p = 0.67$ --- proportion of Good TV

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$SE_n = |p - p_n|$

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$\left[\begin{array}{ccc}TV&Good&SE\\1,2,3,4&2&0.17&2,3,4,5&2&0.17&3,4,5,6&2&0.17\\4,5,6,1&3&0.08&5,6,1,2&4&0.33&6,1,2,3&3&0.08\\1,2,3,5&3&0.08&3,5,6,2&3&0.08&1,3,4,5&2&0.17\\1,3,4,6&2&0.17&1,4,5,2&3&0.08&2,4,6,1&3&0.08\\2,4,6,3&2&0.17&2,4,6,5&3&0.08&3,5,6,1&3&0.08\end{array}\right]$

Read more about sampling distributions at:

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