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babunello [35]
3 years ago
5

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

nutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
A) 1900 J/kg K
B) 2270 J/kg. K
C) 3300 J/kg K
D) 3800 J/kg K
E) 4280 J/kg K
Physics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

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