college.com/course.html?courseld=154322748HeplD=2b3e48e6520860bfd5591538a4a5a27b#10001Search...AAOL Video-Serving the best vi...TripAdvisor34 of 40Calculate the specific heat (J/g °C) for a 18.5-g sample of tin that absorbs 183 J when temperature increases from 35.0 °C to 78.6 °C.V AZΑΣφSH =J/g °CRequest AnswerSubmitPart BCalculate the specific heat (J/g °C) for a 23.0 g sample of a metal that absorbs 669 J when temperature increases from 42.1 °C to92.3 °C.ΑΣ.P PearsonContact UsPrivacy PolicyTerms of UsePermissionsCopyright © 2019 Pearson Education Inc. All rights reserved.8:51 PM12/4/2019CipInsprt scdelefgfgf6144fs&%23back8.4.7.KH.pauN.C.Σ%24%23

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<MC 13 on Temperature, Changes in Energy, and Gases
Item 34
<>
34 of 40
Calculate the specific heat (J/g °C) for a 18.5-g sample of tin that absorbs 183 J when temperature increases from 35.0 °C to 78.6 °C.
V AZ
ΑΣφ
SH =
J/g °C
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Part B
Calculate the specific heat (J/g °C) for a 23.0 g sample of a metal that absorbs 669 J when temperature increases from 42.1 °C to92.3 °C.
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P Pearson
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college.com/course.html?courseld=154322748HeplD=2b3e48e6520860bfd5591538a4a5a27b#10001 Search... A AOL Video-Serving the best vi... TripAdvisor <MC 13 on Temperature, Changes in Energy, and Gases Item 34 <> 34 of 40 Calculate the specific heat (J/g °C) for a 18.5-g sample of tin that absorbs 183 J when temperature increases from 35.0 °C to 78.6 °C. V AZ ΑΣφ SH = J/g °C Request Answer Submit Part B Calculate the specific heat (J/g °C) for a 23.0 g sample of a metal that absorbs 669 J when temperature increases from 42.1 °C to92.3 °C. ΑΣ. P Pearson Contact Us Privacy Policy Terms of Use Permissions Copyright © 2019 Pearson Education Inc. All rights reserved. 8:51 PM 12/4/2019 Cip Ins prt sc dele fg fg f6 144 fs & %23 back 8. 4. 7. K H. pau N. C. Σ %24 %23

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Expert Answer

Step 1

Part A

The specific heat can be calculated as,

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Q= MCAT Q is the amount of heat, C is the specific heat capacity, m is the mass and AT is the difference in the temperature.

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Step 2

Substituting the values

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Q=MCAT 183 = 18.5 x C x (78.6 – 35) C= 0.227 J/g – °C

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Step 3

Part B

The specific heat can be cal...

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Q= MCAT Q is the amount of heat, C is the specific heat capacity, m is the mass and AT is the difference in the temperature.

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