Problem 17.95 A copper calorimeter can with mass 0.443 kg contains 0.0950 kg of

Question

Problem 17.95

A copper calorimeter can with mass 0.443 kg contains 0.0950 kg of ice. The system is initially at 0.0 C.

Part A

If 0.0350 kg of steam at 100.0 C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents?

SubmitMy AnswersGive Up

Part B

At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

Enter your answers numerically separated by commas.

SubmitMy AnswersGive Up

Problem 17.95

A copper calorimeter can with mass 0.443 kg contains 0.0950 kg of ice. The system is initially at 0.0 C.

Part A

If 0.0350 kg of steam at 100.0 C and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents?

T =   C  

SubmitMy AnswersGive Up

Part B

At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

Enter your answers numerically separated by commas.

mice,mwater,msteam =   kg  

SubmitMy AnswersGive Up

Explanation / Answer

(A) energy released by steam as it converts to water at 100 deg.

Q1 = m Lv = 0.0350 x 2260 kJ/kg = 79.1 kJ

energy released as temp down to T from 100 deg C:

Q2 = 0.0350 x 4.186 x (100 - T) = 14.65 - 0.147 T

suppose final temp is T.

Energy absorbed by ice as it converts to waterat 0 deg C,

Q1' = m Lf = 0.0950 x 334 = 31.73 kJ

as temp of water and copper rises to T

Q2' = (0.0950 x 4.186 x T) + (0.443 x 0.385 x T)

Q2' = 0.568T

  
Applying energy conservation assuming no heat loss,

Q1 + Q2 = Q1' + Q2'

79.1 + 14.65 - 0.147T = 31.73 + 0.568T

T = 86.7 deg C .......Ans

(B) m_ice = 0

m_water = 0.0950 + 0.0350 = 0.130 kg

w_steam = 0

Ans: (0, 0.130 kg , 0 )

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---