Specific Heat Review Cp qmt Calculate the amount
- Slides: 7
Specific Heat Review Cp = q/(m)(Δt)
Calculate the amount of heat required to raise the temperature of 78. 2 g 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. 2 g, T 1 = 10 C, T 2 = 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. 2 g)(35 C - 10 C) = 8171. 9 J
What is the specific heat of a 75 gram sample that requires 1200 cal 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 = 75 g, Heat (q) = 1200 cal, T 1 = 25 C, T 2 = 85 F • Cp = ? cal/(g C) • T 2 needs to be converted from F to C: C = 5/9 ( F – 32) => 5/9 (85 – 32) = 29. 4 C = T 2 • Cp = q / (m)( T) Cp = 1200 cal / (75 g)(29. 4 C - 25 C) = 3. 636 cal/(g C)
Calculate the amount of heat, in joules, need to raise 34 g 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)(34 g)(67 C - 55 C) = 852. 72 J
Calculate the specific heat for a 102 g sample that requires 1430 J to raise the temperature from 8. 7 C to 12. 5 C. • • Pick your equation Your right!!!. . Cp= q / (m)( T) Cp= 1430 J / (102 g)(12. 5 C – 8. 7 C) = 3. 69 J/(g C)
• • What will the final temperature be if, at room temperature (25 C), 1300 cal are added to a 76 g sample of iron? Pick your equation Your right!!!. . . . Tf = [q /(m)(Cp)] + Ti Tf= [1300 cal / (76 g)(0. 11 cal/g C)] + 25 C = 180. 5 C
What would the change in temperature ( T) be if an 89 g sample of copper required 678 calories of heat? • • Pick your equation Your right!!!!. . T = q / (m)(Cp) = T = 678 cal/ (89 g)(0. 092 cal/g C) = 82. 8 C
