Question

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

Answer 1

Answer:

b. 16.5°C

Explanation:

$m_{c}$ = mass of piece of copper = 80 g

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

$T_{ci}$ = Initial temperature of piece of copper = 295 °C

$m_{w}$ = mass of water = 250 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of piece of copper = 10 °C

$m_{al}$ = mass of calorimeter  = 300

$c_{al}$ = specific heat of calorimeter = 0.215 cal/g°C

$T_{ali}$ = Initial temperature of calorimeter = 10 °C

$T$ = Final equilibrium temperature

Using conservation of heat

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

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