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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0∘C. Two metallic blocks are placed into the water. One is a 100.0-g piece of copper at 50.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C.
(a) Determine the specific heat of the unknown sample.
(b) Using the data in the table above, can you make a positive identification of the unknown material?
(c) Can you identify a possible material?
(d) Explain your answers for part (b).

Respuesta :

Answer:

a) [tex]c=1822.3214\ J.kg^{-1}.K^{-1}[/tex]

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

  • mass of aluminium, [tex]m_a=0.1\ kg[/tex]
  • mass of water, [tex]m_w=0.25\ kg[/tex]
  • initial temperature of the system, [tex]T_i=10^{\circ}C[/tex]
  • mass of copper block, [tex]m_c=0.1\ kg[/tex]
  • temperature of copper block, [tex]T_c=50^{\circ}C[/tex]
  • mass of the other block, [tex]m=0.07\ kg[/tex]
  • temperature of the other block, [tex]T=100^{\circ}C[/tex]
  • final equilibrium temperature, [tex]T_f=20^{\circ}C[/tex]

We have,

specific heat of aluminium, [tex]c_a=910\ J.kg^{-1}.K^{-1}[/tex]

specific heat of copper, [tex]c_c=390\ J.kg^{-1}.K^{-1}[/tex]

specific heat of water, [tex]c_w=4186\ J.kg^{-1}.K^{-1}[/tex]

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

[tex]Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)[/tex]

[tex]Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)[/tex]

[tex]Q_{in}=11375\ J[/tex]

The heat released by the blocks when dipped into water:

[tex]Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)[/tex]

where

[tex]c=[/tex] specific heat of the unknown material

For the conservation of energy : [tex]Q_{in}=Q_{out}[/tex]

so,

[tex]11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)[/tex]

[tex]c=1822.3214\ J.kg^{-1}.K^{-1}[/tex]

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

onyebuchinnaji

The specific heat capacity of the unknown metal sample is 1.82 J/g⁰C.

The possible material is Beryllium.

The given parameters;

  • mass of the calorimeter, = 100 g
  • mass of water = 250 g
  • temperature of calorimeter, = 10⁰C
  • mass of copper = 100 g
  • temperature of the copper = 50 ⁰C
  • mass of the second metal, = 70 g
  • initial temperature of the second metal = 100 ⁰C
  • final temperature of the entire system, = 20 ⁰C
  • specific of water (4.184 J/g⁰C), aluminum (0.89 J/g⁰C), copper (0.385 J/g⁰C)

Let the specific heat capacity of the unknown sample = [tex]C_x[/tex]

Apply the principle of conservation of energy to determine the specific heat of the unknown sample.

Heat lost by the metals = Heat absorbed by water and calorimeter

[tex]mc \Delta \theta _{hot} = mc \Delta \theta _{cold} \\\\100 (0.385)(50 - 20) + 70C_x(100 - 20) = [100 ( 0.89) \ + \ 250(4.184 )](20-10)\\\\1155 + 5600C_x = 11350\\\\5600C_x = 10195\\\\C_x = \frac{10195}{5600} \\\\C_x = 1.82 \ J/g^0 C[/tex]

The specific heat capacity of the unknown metal sample corresponds  Beryllium. Thus, we can conclude that the possible material is Beryllium.

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