Question

A hot lump of 32.3g of Copper at an initial temperature of 96.5 degrees Celsius in 50mL H2O initially at 25.0 degrees Celsius and allowed to reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific hear is 0.385J/g°C and the specific heat of water is 4.184J/g°C?

4. A hot lump of 41.3 g of copper at an initial temperature of 94.8 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/g°C and the specific heat of water is 4.184 J/g°C? b. 27.4°C a. 25.0°C c. 28.3°C d. 29.9°C

Answer 1

Given :

Mass of copper = 41.3 g

Specific heat capacity of copper = 0.385 J / g ⁰ C

Initial temperature of copper =94.8 ⁰ C

Volume of water = 50.0 ml

Initial temperature of water = 25.0 ⁰ C

Specific heat capacity of water = 4.184 J / g ⁰ C

final temperature of copper = ?

final temperature of water = ?

Initial temperature of metal is higher than initial temperature of water. When metal is added to water, heat will flow from metal to water i e temperature of metal will decrease and that of water will increase.

Hence, we can write Heat lost by metal = Heat gain by water

i e q copper = - ( q water )

We have relation, q = m $\times$ C $\times$ $\Delta$T

Where, q is amount of heat absorbed or released by an substance, m is mass of substance, C is specific heat capacity of substance & $\Delta$ T is temperature change of substance.

$\therefore$ [ m $\times$ C $\times$ $\Delta$T ] copper = - [ m $\times$ C $\times$ $\Delta$T ] water

41.3 g $\times$ 0.385 J / g ⁰ C  $\times$ ( T final - 94.8 ⁰ C ) = - [ m $\times$ 4.184 J / g ⁰ C  $\times$ ( T final -25.0 ⁰ C ) ]

We have relation, density = Mass / volume

$\therefore$ Mass of substance = density $\times$ volume = 0.997 g / ml $\times$ 50.0 ml = 49.85 g

$\therefore$ 41.3 g $\times$ 0.385 J / g ⁰ C  $\times$ ( T final - 94.8 ⁰ C ) = - [49.85 g  $\times$ 4.184 J / g ⁰ C  $\times$ ( T final -25.0 ⁰ C ) ]

15.9 J / ⁰ C $\times$ ( T final - 94.8 ⁰ C ) = - [ 208.6 J / ⁰ C  $\times$ ( T final -25.0 ⁰ C ) ]

$\therefore$  15.9 J / ⁰ C $\times$   T final - 1507 J = - 208.6 J / ⁰ C  $\times$ T final + 5218 J

$\therefore$  15.9 J / ⁰ C $\times$   T final + 208.6 J / ⁰ C  $\times$ T final = 5218 J + 1507 J  

224.5 J / ⁰ C $\times$   T final = 6725 J

   T final = 6725 J / ( 224.5 J / ⁰ C )

     T final = 29.96  ⁰ C

ANSWER : d ) 29.9  ⁰ C