Question

A 286.0 g piece of granite, heated to 571.0°C in a campfire, is dropped into 1.15 L water (d = 1.00 g/mL) at 25.0°C. The molar heat capacity of water is cp,water = 75.3 J/(mol ·°C), and the specific heat of granite is cs,granite = 0.790 J/(g ·°C).

Calculate the final temperature of the granite in Celcius

Answer 1

Given

Mass of granite = 286.0 g

Specific heat capacity of granite = 0.790 J / g ⁰ C

Initial temperature of granite = 571.0 ⁰ C

Volume of water = 1.15 L = 1150 ml

Density of water = 1.00 g / ml

Molar heat capacity of water = 75.3 J / mol ⁰ C

Initial temperature of water = 25.0 ⁰ C

We can solve given problem in following steps.

Step 1 : Calculation of mass of water and no. of moles of water

We have formula, density = Mass / volume

Mass of water = density volume

Mass of water = 1.00 g /ml 1150 ml = 1150 g

We have, mo. of moles = Mass / Molar mass

No. of moles of water = 1150 g / ( 18.02 g /mol ) = 63.82 mol

Step 2 : Calculation of equilibrium temperature.

Assume no heat is lost to surrounding. Therefore, Heat lost by metal is absorbed by water.Hence, we can write

q metal + q water = 0

Heat absorbed or emitted by any substance is given as q = m C (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.

( m C (T final - T initial ) ) metal +  ( m C (T final - T initial ) ) water = 0

Substituting given values in above equation , we get

( 286.0 g 0.790 J / g ⁰ C (T final - 571.0 ⁰ C ) + ( 63.82 mol 75.3 J / mol ⁰ C ( (T final - 25.0 ⁰ C )) = 0

225.94 J / ⁰ C (T final - 571.0 ⁰ C ) + 4805.65  J / ⁰ C  ( (T final - 25.0 ⁰ C )) = 0

225.94 J / ⁰ C T final - 129011.7 J + 4805.65  J / ⁰ C T final - 120141.25 J = 0

225.94 J / ⁰ C T final+ 4805.65  J / ⁰ C T final = 249152.95 J

5031.59 J / ⁰ C T final = 249152.95 J

T final = 249152.95 J / 5031.59 J / ⁰ C

T final = 49.5  ⁰ C

ANSWER : Final temperature of water = 49.5  ⁰ C