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# 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

## Specific Heat Review

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### 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

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