Specific Heat Review Cp = q/(m)(Δt).

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Presentation transcript:

Specific Heat Review Cp = q/(m)(Δt)

Calculate the amount of heat required to raise the temperature of 78 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

Manipulate the formula to solve for specific heat: Cp = q/ (m)(T) 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)

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

Cp= 1430J / (102g)(12.5C – 8.7C) = 3.69 J/(gC) 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)

Tf= [1300cal / (76g)(0.11 cal/gC)] + 25C = 180.5C 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

T = 678 cal/ (89g)(0.092 cal/gC) = 82.8C 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