Calorimetry An aluminum container with a mass of 500 g contains 500 g of water a

Question

Calorimetry

An aluminum container with a mass of 500 g contains 500 g of water at 100 0 C. A 1.00 kg block of copper at 0.00 0 C is then dropped into the container.

(a) What is the final temperature of the system? Assume an isolated system.

(b) How much internal energy was transferred to the block? What is the block’s change in temperature?

(c) How much internal energy did the aluminum container transfer to the block? What is the container’s change in temperature?

(d) How much internal energy did the water transfer to the block? What is the water’s change in temperature?

(e) The water gives up much more internal energy than the container but they both have the same mass and the same temperature change. Explain why

Explanation / Answer

we know,

C_aluminmum = 900 J/(kg C)

C_copper = 380 J/(kg C)

C_water = 4186 J/(kg C)

a) let T is the final equilibrium temperature

heat lost by aluminum and water = heat gained by copper

m_water*C_water*(100 - T) + m_aluminum*C_aluminum*(100 - T) = m_copper*C_copper*(T - 0 )

0.5*4186*(100 - T) + 0.5*900*(100 -T) = 1*380*(T - 0 )

==> T = 87 degrees celsius

b) energy transfred to the block = m_copper*C_copper*(T - 0 )

= 1*380*(87 - 0 )

= 33060 J

c) Energy tranefred to the block from alunimum = m_aluminum*C_aluminum*(100 - T)

= 0.5*900*(100 -87)

= 5850 J

change in temperature of aluminum = 100 - 87

= 13 degrees clesius

d) Energy tranefred to the block from water = m_water*C_water*(100 - T)

= 0.5*4186*(100 - 87)

= 27209 J

change in temperature of water = 100 - 87

= 13 degrees clesius

e) Because, water has more specific heat than aluminum.

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