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jolli1 [7]
3 years ago
13

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).
Physics
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

a) c=1822.3214\ J.kg^{-1}.K^{-1}

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

We have,

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

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

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

Using the heat energy conservation equation.

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

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

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

Q_{in}=11375\ J

The heat released by the blocks when dipped into water:

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

where

c= specific heat of the unknown material

For the conservation of energy : Q_{in}=Q_{out}

so,

11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)

c=1822.3214\ J.kg^{-1}.K^{-1}

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.

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Given that R_{1} = 200\; {\Omega} and R_{2} = 250\; {\Omega}:

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