Question

An irregular lump of an unknown metal has a measured density of 5.92 g/mL. The metal is heated to a temperature of 191 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.3 mL, and the temperature is recorded as 40.8 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Answer 1

Given

Density of metal = 5.92 g /ml

Initial temperature of metal = 191 ⁰ C

Initial Volume of water = 25.0 ml

Volume of water after addition of metal = 34.3 ml

Initial temperature of water = 25 ⁰ C

Final temperature of water= 40.8 ⁰ C

Final temperature of metal = 40.8 ⁰ C

First we have to calculate mass of metal.

From given information, volume of metal = Volume of water after addition of metal - Initial Volume of water

$\therefore$ volume of metal = 34.3 ml - 25.0 ml = 9.3 ml

We know that, density = Mass / volume

$\therefore$ Mass of metal = density $\times$ volume = 5.92 g /ml $\times$ 9.3 ml = 55.056 g

Similarly, Mass of water = 0.99707 g /ml $\times$ 25.0 ml = 24.93 g ( density of water at 25 ⁰ C = 0.99707 g/ml)

In this case, Heat lost by metal is absorbed by water.No heat is transferred to surrounding.

Hence, we can write q _metal + q _water = 0

Heat absorbed or emitted by any substance is given as q = m x C x (T final - T initial )

Where q is a heat absorbed or emitted, m is a mass of a body, C is a specific heat capacity of a body, T is a temperature of a body.

$\therefore$ [ 55.056 g $\times$ C $\times$ ( 40.8 ⁰ C - 191 ⁰ C)] _metal + [ 24.93 g $\times$ 4.184 J / g ⁰ C $\times$ ( 40.8 ⁰ C - 25.0 ⁰ C)] _water = 0

[ 55.056 g $\times$ C $\times$ (- 150.2  ⁰ C )] _metal + 1648.0 J = 0

[ 55.056 g $\times$ C $\times$ (- 150.2  ⁰ C )] _metal = - 1648.0 J

C metal = - 1648.0 J / ( 55.056 g $\times$ (- 150.2  ⁰ C ) = 1648.0 J / 8269.4 g ⁰ C = 0.199 J / g ⁰ C

ANSWER : C metal = 0.199 J / g ⁰ C