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

0.167 J/(g °C)

Step-by-step 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)