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1)Calculate ∆T for 4.0 gram water absorbing 33 J. q = mC ∆T 33 J = 4.0g * 4.18 J * ∆T o C

1)Calculate ∆T for 4.0 gram water absorbing 33 J. q = mC ∆T 33 J = 4.0g * 4.18 J * ∆T o C g o C ∆T = 2.02…choice 2. ∆T is positive when T increases, and q is positive. 2) Calculate T 2 for 2.0 grams of water at 15 o C absorbing 84 J. q = mC(T 2 – T 1 )

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1)Calculate ∆T for 4.0 gram water absorbing 33 J. q = mC ∆T 33 J = 4.0g * 4.18 J * ∆T o C

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  1. 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 positivewhen 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

  2. 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

  3. 5) Calculate J absorbedby 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

  4. 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

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