1 / 7

Specific Heat Review

Specific Heat Review. C p = q/(m)( Δ t). Calculate the amount of heat required to raise the temperature of 78.2g of water from 10  C to 35  C. Manipulate the formula to solve for heat: C p = q solve for q = (C p )(m)(  T) m Δ t

kynton
Télécharger la présentation

Specific Heat Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Specific Heat Review Cp = q/(m)(Δt)

  2. Calculate the amount of heat required to raise the temperature of 78.2g of water from 10C to 35C. • Manipulate the formula to solve for heat: • Cp= q solve for q = (Cp)(m)(T) m Δt • Check the units: m = 78.2g, T1 = 10C, T2 = 35C • Cp is not given, but we are dealing with water so Cp = 4.18 J/(gC) • Heat = ? J • All the units match-up, so we do not have to convert any units prior to plugging into the equation. • Heat = Cp(m)(T) Heat = [4.18 J/(gC)] (78.2g)(35C - 10C) = 8171.9 J

  3. What is the specific heat of a 75 gram sample that requires 1200cal to change the temperature from 25C to 85F? • Manipulate the formula to solve for specific heat: Cp = q/ (m)(T) • Check the units: m = 75g, Heat (q) = 1200cal, T1 = 25C, T2 = 85F • Cp = ? cal/(gC) • T2 needs to be converted from F to C: C = 5/9 (F – 32) => 5/9 (85 – 32) = 29.4C = T2 • Cp = q / (m)(T) Cp = 1200cal / (75g)(29.4C - 25C) = 3.636 cal/(gC)

  4. Calculate the amount of heat, in joules, need to raise 34g of ice (Cpice= 2.09 J/gC) from 55C to 67C. • Pick your equation • Your right!!!!... • q = Cp(m)(T) = • (2.09 J/gC)(34g)(67C - 55C) = 852.72J

  5. Calculate the specific heat for a 102g sample that requires 1430J to raise the temperature from 8.7C to 12.5C. • Pick your equation • Your right!!!.. • Cp= q / (m)(T) • Cp= 1430J / (102g)(12.5C – 8.7C) = 3.69 J/(gC)

  6. What will the final temperature be if, at room temperature (25C), 1300 cal are added to a 76g sample of iron? • Pick your equation • Your right!!!........ • Tf = [q /(m)(Cp)] + Ti • Tf= [1300cal / (76g)(0.11 cal/gC)] + 25C = 180.5C

  7. What would the change in temperature (T) be if an 89g sample of copper required 678 calories of heat? • Pick your equation • Your right!!!!.... • T = q / (m)(Cp) = • T = 678 cal/ (89g)(0.092 cal/gC) = 82.8C

More Related