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An 80.0-g piece of copper, initially at 295°C, is dropped into 250 g of water contained in a 300-g aluminum calorimeter; the water and calorimeter are initially at 10.0°C.

What is the final temperature of the system? (Specific heats of copper and aluminum are 0.092 0 and 0.215 cal/g⋅°C, respectively. cw = 1.00 cal/g°C)
a. 12.8°C
b. 16.5°C
c. 28.4°C
d. 32.1°C

Respuesta :

Answer:

b. 16.5°C

Explanation:

[tex]m_{c}[/tex] = mass of piece of copper = 80 g

[tex]c_{c}[/tex] = specific heat of piece of copper = 0.0920 cal/g°C

[tex]T_{ci}[/tex] = Initial temperature of piece of copper = 295 °C

[tex]m_{w}[/tex] = mass of water = 250 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/g°C

[tex]T_{wi}[/tex] = Initial temperature of piece of copper = 10 °C

[tex]m_{al}[/tex] = mass of calorimeter  = 300

[tex]c_{al}[/tex] = specific heat of calorimeter = 0.215 cal/g°C

[tex]T_{ali}[/tex] = Initial temperature of calorimeter = 10 °C

[tex]T[/tex] = Final equilibrium temperature

Using conservation of heat

Heat lost by piece of copper = heat gained by water + heat gained by calorimeter

[tex]m_{c} c_{c} (T_{ci} - T) = m_{w} c_{w} (T - T_{wi})+ m_{al} c_{al} (T - T_{ali})\\(80) (0.092) (295 - T) = (250) (1) (T - 10) + (300) (0.215) (T - 10)\\T = 16.5 C[/tex]

onyebuchinnaji

The final temperature of the system is 16.4 ⁰C.

Conservation of energy

The final temperature of the system is determined by applying the principle of conservation of energy as shown below;

Heat lost by piece of copper = heat gained by water + heat gained by calorimeter

mCc(Tc - T) = mCw(T - Tw) + mCl(T - Tl)

80 x 0.09 x (295 - T) = 250 x 1 x (T - 10) + 300 x 0.215 x (T - 10)

2124 - 7.2T = 250T - 2500 + 64.5T - 645

5269 = 321.7T

T = 5269/321.7

T = 16.4 ⁰C

Thus, the final temperature of the system is 16.4 ⁰C.

Learn more about conservation of energy here: https://brainly.com/question/166559

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