spurlingl
Apr 15, 2010, 12:54 AM
basically, I would like it if you could just check my work...
1. Calculate the quantity of heat required to raise 20.1g of lead from 5.0 C to 45 C. The specific heat of lead is 0.128 J/g*C.
Q= 20.1g x .128 J/g*C x [45-5]
Q= 102.912 J correct?
__________________________________________________ _______________________
2. Calculate the quantity of heat lost when 52g of S is cooled from 85 C to 60 C. The specific heat of sulfur is 0.736 J/g*C.
Q= 52g x .736 J/g*C x [60-80]
Q= -956.8 J yes?
__________________________________________________ _______________________
3. In cooling from 75 C to 43 C, 25g of a certain substance are found to lose 669.5 J. Calculate the specific heat of the substance.
-669.5 J = 25g x c x [43-75]
-699.5/(25 x -32) = c
c = .836875 J/g*C right?
__________________________________________________ _______________________
4. A 46g sample of iron lost 297 J of heat when it cooled down to 67 C. If the specific heat of iron is 0.460 J/g*C, calculate the initial temperature.
-297 J = 46g x .460 J/g*C x [67-T(I)]
-297 = 21.16 x [67-T(I)]
-297 = 1417.72 - 21.16T(I)
-2011.72 = -21.16T(I)
-2011.72/-21.16 = T(I)
T(I) = 95 C ?
thank you so much!
1. Calculate the quantity of heat required to raise 20.1g of lead from 5.0 C to 45 C. The specific heat of lead is 0.128 J/g*C.
Q= 20.1g x .128 J/g*C x [45-5]
Q= 102.912 J correct?
__________________________________________________ _______________________
2. Calculate the quantity of heat lost when 52g of S is cooled from 85 C to 60 C. The specific heat of sulfur is 0.736 J/g*C.
Q= 52g x .736 J/g*C x [60-80]
Q= -956.8 J yes?
__________________________________________________ _______________________
3. In cooling from 75 C to 43 C, 25g of a certain substance are found to lose 669.5 J. Calculate the specific heat of the substance.
-669.5 J = 25g x c x [43-75]
-699.5/(25 x -32) = c
c = .836875 J/g*C right?
__________________________________________________ _______________________
4. A 46g sample of iron lost 297 J of heat when it cooled down to 67 C. If the specific heat of iron is 0.460 J/g*C, calculate the initial temperature.
-297 J = 46g x .460 J/g*C x [67-T(I)]
-297 = 21.16 x [67-T(I)]
-297 = 1417.72 - 21.16T(I)
-2011.72 = -21.16T(I)
-2011.72/-21.16 = T(I)
T(I) = 95 C ?
thank you so much!