Question

Case 2: A 0.780-kg silver pellet with a temperature of 85 oC is added to 0.150 kg of water in a copper cup of 0.66 kg. The initial temperature of the water and the copper cup is 14 oc. Assume no heat is exchanged with the surroundings. The specific heats of silver, water and copper are: 234 J/(kg oC), 4186 J/(kg oC) and 387 J/(kg oc), respectively. Find the final temprature of the system (silver+water+copper cup). Note: 1. Slver pellet releases heat, apply Q mcAT to silver. To ensure Q is positive, how should you write AT for silver, in terms of the final temperature Tf and silver's intial temperature? (no submission) 2. Water absorbs heat, apply Q mcAT to water. To ensure Q is positive, how should you write AT for water, in terms of the final temperature Tf and water's intial temperature? (no submission) 3. Copper cup absorbs heat, apply Q mcaT to copper. To ensure Qis positive, how should you write AT for copper, in terms of the final temperature Tf and copper's intial temperatu re?? (no submission) 3. Let Q released by silver Q absorbed by water + Q absorbed by copper solve for Tf Check that your result of Ty is reasonable. Keep 2 decimal places. (5 attempts remaining) Case 3: A 0.085 kg of an unknown solid sample at a temperature of 100.0°C is dropped into a calorimetor. The calorimetor can is made of 0.56 kg of copper (specific heat 387 J(kg °C)) and contains 0.150 kg of water (specific heat 4186 J/kg oC)), and both the can and water are initially at 14.0°C. The final temprature of the system after thermal equilibrium is established is measured to be 22.5 °C. Assume no heat is exchanged with the surroundings. Calculate the heat absorbed by water. Keep 2 decimal places. Report all heat as POSITIVE (5 attempts remaining) Calculate the heat absorbed by copper. Keep 2 decimal places. Report all heat as POSITIVE (5 attempts remaining) Find the specific heat (in J/(kg oC) of this unknown sample, Keep 1 decimal place. JI(kgoC) (5 attempts remaining)

Answer 1

For silver, Qs = 0.78*234*(85-Tf)

For copper, Qc = 0.66*387*(Tf-14)

For water, Qw = 0.15*4186*(Tf-14)

Qs = 0.78*234*(85-Tf) = Qc + Qw = 0.66*387*(Tf-14) + 0.15*4186*(Tf-14)
15514.2 - 182.52 Tf = 255.42 Tf - 3575.88 + 627.9 Tf - 8790.6
Tf = 26.15 C

Case 3:

For sample, Qs = 0.085*c*(100-22.5)

For copper, Qc = 0.56*387*(22.5-14) = 1842.12

For water, Qw = 0.15*4186*(22.5-14) = 5337.15

Qs = 0.085*c*(77.5) = Qc + Qw = 0.56*387*(8.5) + 0.15*4186*(8.5)
6.5875c= 7179.27
c = 1089.8 J/kg C