5. A student wishes to determine the heat capacity ofa coffee-cup calorimeter. A

Question

5.

A student wishes to determine the heat capacity ofa coffee-cup calorimeter. After she mixes 100.0 g of water at 58.5degrees C with 100.0 g of water, already in the calorimeter, at22.8 degrees C, the final temperature of the water is 39.7 degreesC. Calculate the heat capacity of the calorimeter in J/degrees C.(Use 4.184 J/g*degrees C as the specific heat of water.)
Student Response Value Correct Answer Answer: not answered   0% 47 General Feedback: Calorimetry is covered in Section 6.8 of the text. Score: 0/1

6.

A reaction occurs in a calorimeter that contains222.0 g of water at 22.49 oC, and 5.60 kJ of heat isevolved. If the specific heat capacity of water is 4.184J/oC*g, what is the final temperature of the water ifthe calorimeter does not absorb any of the heat?
Student Response Value Correct Answer Answer: not answered   0% 28.52 Units: not answered 0.0% °C General Feedback: Section 6.8 covers calorimetry. Rearrange the heat equation toisolate DT. Then, remember to add thechange in temperature to the intial temperature. Score: 0/1

5.

A student wishes to determine the heat capacity ofa coffee-cup calorimeter. After she mixes 100.0 g of water at 58.5degrees C with 100.0 g of water, already in the calorimeter, at22.8 degrees C, the final temperature of the water is 39.7 degreesC. Calculate the heat capacity of the calorimeter in J/degrees C.(Use 4.184 J/g*degrees C as the specific heat of water.)
Student Response Value Correct Answer Answer: not answered   0% 47 General Feedback: Calorimetry is covered in Section 6.8 of the text. Score: 0/1

6.

A reaction occurs in a calorimeter that contains222.0 g of water at 22.49 oC, and 5.60 kJ of heat isevolved. If the specific heat capacity of water is 4.184J/oC*g, what is the final temperature of the water ifthe calorimeter does not absorb any of the heat?
Student Response Value Correct Answer Answer: not answered   0% 28.52 Units: not answered 0.0% °C General Feedback: Section 6.8 covers calorimetry. Rearrange the heat equation toisolate DT. Then, remember to add thechange in temperature to the intial temperature. Score: 0/1

Explanation / Answer

5. Heat = mass * specific heat * temperature difference Heat absorbed by calorimeter = Heat capacity *temperature difference . Heat given out by water at 58.5 C = Heat absorbed by water at22.8 + calorimeter at 22.8 C 100 g * 4.18 J/(g.C) * (58.5 - 39.7) = 100 g * 4.18 J/(g. C) *(39.7 - 22.8) C + Heat capacity * (39.7-22.8) Solving, we get Heat capacity of the calorimeter = 46.99J/C . 6 Let final temperature = Tf Heat = mass * specific heat * temperature difference 5.60 * 1000 J = 222 g * 4.184 J/ (C.g) * ( Tf - 22.49) Solving, we get Tf = 28.52 C

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