Question

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 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.

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

Answer 1

2274 J/kg ∙ K

Step-by-step 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)`