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a chemist carefully measures the amount of heat needed to raise the tem…

Question

a chemist carefully measures the amount of heat needed to raise the temperature of a 301.0 mg sample of a pure substance from 46.2 ^ { circ } c to 60.1 ^ { circ } c. the experiment shows that 17.5 j of heat are needed. what can the chemist report for the specific heat capacity of the substance? be sure your answer has the correct number of significant digits.

Explanation:

Step1: Recall the formula for heat

The formula relating heat (\(q\)), mass (\(m\)), specific heat capacity (\(c\)), and temperature change (\(\Delta T\)) is \(q = mc\Delta T\). We need to solve for \(c\), so \(c=\frac{q}{m\Delta T}\).

Step2: Convert mass to grams

The mass \(m = 301.0\space mg\). Since \(1\space g=1000\space mg\), we convert: \(m=\frac{301.0\space mg}{1000\space mg/g}=0.3010\space g\).

Step3: Calculate temperature change

The initial temperature \(T_1 = 46.2^\circ C\), final temperature \(T_2 = 60.1^\circ C\). The temperature change \(\Delta T=T_2 - T_1=60.1 - 46.2 = 13.9\space K\) (since the change in Celsius is the same as the change in Kelvin).

Step4: Substitute values into the formula

We know \(q = 17.5\space J\), \(m = 0.3010\space g\), \(\Delta T = 13.9\space K\). Substitute into \(c=\frac{q}{m\Delta T}\):

$$c=\frac{17.5\space J}{0.3010\space g\times13.9\space K}$$

Step5: Calculate the value

First, calculate the denominator: \(0.3010\times13.9 = 4.1839\)
Then, \(c=\frac{17.5}{4.1839}\approx4.18\space J\space g^{-1}\space K^{-1}\) (we check significant digits: \(q = 17.5\) (3 sig figs), \(m = 301.0\) (4 sig figs), \(\Delta T=13.9\) (3 sig figs). The least number of sig figs in multiplication/division is 3, so our answer should have 3 sig figs. Wait, wait, let's recalculate:

Wait, \(0.3010\times13.9\): \(0.3010\) has 4 sig figs, \(13.9\) has 3. So the product has 3 sig figs: \(0.3010\times13.9 = 4.1839\approx4.18\) (3 sig figs). Then \(17.5\div4.18\approx4.186\)? Wait, no, let's do the division properly:

\(17.5\div(0.3010\times13.9)=17.5\div4.1839\approx4.18\)? Wait, no, \(0.3010\times13.9 = 0.3010\times13 + 0.3010\times0.9=3.913+0.2709 = 4.1839\). Then \(17.5\div4.1839\approx4.18\) (wait, 17.5 divided by 4.1839: 4.1839*4 = 16.7356, 17.5 - 16.7356 = 0.7644, 0.7644/4.1839≈0.1827, so total≈4.1827, which rounds to 4.18 with 3 sig figs? Wait, but 17.5 has 3, 0.3010 has 4, 13.9 has 3. So the rule is that in multiplication/division, the result has the same number of sig figs as the least precise measurement. Here, 17.5 (3), 13.9 (3), so the result should have 3 sig figs. Wait, but let's check again:

Wait, the mass is 301.0 mg, which is 0.3010 g (4 sig figs), temperature change is 13.9 K (3 sig figs), heat is 17.5 J (3 sig figs). So when we do \(q/(m\Delta T)\), the number of sig figs is determined by the least, which is 3 (from q and \(\Delta T\)). So let's recalculate:

\(m = 301.0\space mg = 0.3010\space g\) (4 sig figs)

\(\Delta T = 60.1 - 46.2 = 13.9\space ^\circ C = 13.9\space K\) (3 sig figs)

\(q = 17.5\space J\) (3 sig figs)

So \(c=\frac{17.5}{0.3010\times13.9}=\frac{17.5}{4.1839}\approx4.18\space J\space g^{-1}\space K^{-1}\)? Wait, no, wait 17.5 divided by 4.1839: 4.1839*4 = 16.7356, 17.5 - 16.7356 = 0.7644, 0.7644/4.1839≈0.1827, so 4.1827, which is approximately 4.18 when rounded to three significant figures? Wait, but let's check the calculation again. Wait, maybe I made a mistake in the temperature change. Wait, 60.1 - 46.2 is 13.9, correct. Mass: 301.0 mg is 0.3010 g, correct. Heat: 17.5 J. So:

\(c=\frac{17.5\space J}{0.3010\space g\times13.9\space K}=\frac{17.5}{4.1839}\approx4.18\space J\space g^{-1}\space K^{-1}\)? Wait, no, 17.5 divided by 4.1839 is approximately 4.18? Wait, 4.18394.18 = 4.18394 + 4.1839*0.18 = 16.7356 + 0.7531 = 17.4887, which is very close to 17.5. So yes, approximately 4.18. Wait, but let's check the significant figures. The mass is 301.0 (4 sig figs), heat is 17.5 (3 sig figs), temperature change is 13.9 (3 sig figs). So the limiting factor is 3 sig figs,…

Answer:

\(4.18\)