1) 70.0 grams of Gold at 100.0 °C is added to 20.0 grams of water at 20.0°C in a

Question

1) 70.0 grams of Gold at 100.0 °C is added to 20.0 grams of water at 20.0°C in an insulated container. The final temparature of he mixture is 33.0 °C. The specific heat of water is 4.184 j/g °C. What is the specific heat of gold?

2) 70.0 grams of Aluminium at 100.0 °C is added to 20.0 grams of water at 20.0°C in an insulated container. The final temparature of he mixture is 33.0 °C. The specific heat of water is 4.184 j/g °C. What is the specific heat of Aluminium?

3) 70.0 grams of Zinc at 100.0 °C is added to 20.0 grams of water at 20.0°C in an insulated container. The final temparature of he mixture is 33.0 °C. The specific heat of water is 4.184 j/g °C. What is the specific heat of Zinc ?

Explanation / Answer

1) 70.0 grams of Gold at 100.0 °C is added to 20.0 grams of water at 20.0°C in an insulated container. The final temparature of he mixture is 33.0 °C. The specific heat of water is 4.184 j/g °C. What is the specific heat of gold?

Solution :-   Metal temperature is higher than the temperature of the water therefore using the following set up we can calculate the specific heat of the metal as follows

-q hot gold = q cold water

-m*c*delta T = m*c*deltaT

- 70.0 g * c* (33.0 C- 100 C) = 20.0 g * 4.184 J per g C* (33.0 C – 20.0 C)

4690 g.oC * c = 1087.84 J

c= 1087.84 J/ 4690 g.oC

c= 0.232 J per g C

Therefore the specific heat of the Gold = 0.232 J per g .oC

2) 70.0 grams of Aluminium at 100.0 °C is added to 20.0 grams of water at 20.0°C in an insulated container. The final temparature of he mixture is 33.0 °C. The specific heat of water is 4.184 j/g °C. What is the specific heat of Aluminium?

Solution :-

-q hot Al = q cold water

-m*c*delta T = m*c*deltaT

- 70.0 g * c* (33.0 C- 100 C) = 20.0 g * 4.184 J per g C* (33.0 C – 20.0 C)

4690 g.oC * c = 1087.84 J

c= 1087.84 J/ 4690 g.oC

c= 0.232 J per g C

Therefore the specific heat of the Aluminium = 0.232 J per g .oC

3) 70.0 grams of Zinc at 100.0 °C is added to 20.0 grams of water at 20.0°C in an insulated container. The final temparature of he mixture is 33.0 °C. The specific heat of water is 4.184 j/g °C. What is the specific heat of Zinc ?

Solution :-

-q hot zinc = q cold water

-m*c*delta T = m*c*deltaT

- 70.0 g * c* (33.0 C- 100 C) = 20.0 g * 4.184 J per g C* (33.0 C – 20.0 C)

4690 g.oC * c = 1087.84 J

c= 1087.84 J/ 4690 g.oC

c= 0.232 J per g C

Therefore the specific heat of the Zinc = 0.232 J per g .oC

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