Transcript 583 groupwork 1-29-10

1)Calculate ∆T for 4.0 gram water absorbing 33 J.
q = mC∆T
33 J = 4.0g * 4.18 J * ∆T oC
goC
∆T = 2.02…choice 2
∆T is positive when
T increases, and q
is positive
2) Calculate T2 for 2.0 grams of water at 15oC
absorbing 84 J.
q = mC(T2 – T1)
84 J = 2.0 g * 4.18 J/ g oC (T2 - 15oC)
T2 = 25oC
3) 50. grams of water is heated, temp increases to 50
oC and 4,180 J of heat is added. What is initial temp?
q = mC∆T, q=mC(T2 – T1)
4180 = 50 g * 4.18 oC/g * ( 50 oC – T1 )
T1= 30 oC
4) 7.00 grams of water is heated, temp increases from
10.0 to 15.0 oC how much energy is absorbed?
q = mC∆T, q=mC(T2 – T1)
q = 7.00 g * 4.18 oC/g * ( 15 oC – 10oC )
q= 146 J
5) Calculate J absorbed by 200.0 g when the temp
increases form 10.0 to 40.0 oC.
q = mC∆T
q = 200.0g * 4.18 * (40.0-10.0)
q = 200.0g * 4.18 * (30 oC) = 25,080 J
6) Find kJ of heat absorbed when 70.0 g water is
vaporized at its boiling pt.
q = mHvap
q = 70.0g * 2260.0 J/g = 158200 J / 1000 = 158.2 kJ
7) How many grams of water absorb 2510 J of energy
when the temp changes from 10 to 30 oC.
q = mC∆T
2510J = X g * 4.18 J * 20oC = 30.06 grams water
10) The temp of 50g of water is raised to 50 oC be the
addition of 1000 cal of energy, find initial temp (T1).
q = mC∆T
1000 = 50g * 1.0 cal g * (50 - T1)
T1 = 30