1. A 88.3 g sample of metal at 95.24 o C is added to 35.10 g of water that is in

Question

1. A 88.3 g sample of metal at 95.24 oC is added to 35.10 g of water that is initially at 17.27 oC. The final temperature of both the water and the metal is 29.20oC. The specific heat of water is 4.184 J/(goC). Calculate the specific heat of the metal.

2. A 56.4 g sample of Al at 93.3 oC is added to 173.7 g of water that is initially at 27.0 oC. The specific heat of Al and water are 0.90 J/(goC) and 4.184 J/(goC), respectively. Calculate the equilibrium temperature.

A 66 mL solution of a dilute AgNO3 solution is added to 88 mL of a base solution in a coffee-cup calorimeter. As AgOH (s) precipitates, the temperature of the solution increases from 23.35 oC to 25.65 oC. Assuming the mixture has the same specific heat (4.184J/goC) and density (1.00 g/cm3 or 1.00 g/mL) as water, calculate the heat (in J) transferred to the surroundings, qsurr.

Explanation / Answer

1)

Le tthe specific heat cap of metal be c..

88.3 * c * ( 95.24 - 29.2 ) = 4.184 * 35.1 * ( 29.2 - 17.27 )

so c = 0.30045 J / (goC )

2)

Let the temp be T

56.4 * 0.9 * ( 93.3 - T ) = 173.7 * 4.184 * ( T - 27 )

4,735.908 - 50.76* T = 726.7608T - 19,622.5416

so T = 31.33 celcius

3)

volume of solution = 88 + 66 = 154 ml

mass of solution = 154 * 1 = 154 g

heat transferred = 154 * 4.184 * ( 25.65 - 23.35 ) = 1,481.9728 Joules

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.