Question

 Problem 1.5 A 250 g sample at 80°C is placed in a calorimeter containing 300 g of water. The mass of the aluminum calorimeter cup is 100 g. The initial temperature of the water and cup is 20.0°C, and the final temperature is 21.8°C. What is the specific heat of the sample?

Answer 1

Answer:

0.167 J/(g °C)

Explanation:

$m_{s}$ = mass of the sample = 250 g

$T_{si}$ = Initial temperature of sample = 80 °C

$m_{w}$ = mass of the water = 300 g

$m_{c}$ = mass of the cup = 100 g

$c_{w}$ = Specific heat of water = 4.186 J/(g °C)

$c_{s}$ = Specific heat of sample = ?

$c_{c}$ = Specific heat of aluminum cup = 0.9  J/(g °C)

$T_{wi}$ = Initial temperature of water = 20 °C

$T_{ci}$ = Initial temperature of aluminum cup = 20 °C

$T_{f}$ = Final temperature = 21.8 °C

Using conservation of heat

Heat lost by sample = Heat gained by water + Heat gained by cup

$m_{s} c_{s} (T_{si} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci})$

$(250) c_{s} (80 - 21.8) = (300) (4.186) (21.8 - 20) + (100) (0.9) (21.8 - 20)$

$c_{s}$ = 0.167 J/(g °C)