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2 months ago

6

cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge wit

h an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.

1 answer:

grin007 [323]2 months ago

3 0

Answer:

final temperature = 26.5°

Explanation:

The water's initial volume is calculated as 1 x 1 x 1 = 1 $m^{3}$

The starting temperature of the water is 20° C

Water density = 1000 kg/ $m^{3}$

Volume of the copper block is determined as 0.46 x 0.46 x 0.46 = 0.097 $m^{3}$

The copper block's initial temperature is 100° C

Copper density = 8960 kg/ $m^{3}$

The water's final volume after accounting for the copper block is 1 - 0.097 = 0.903 $m^{3}$

Assumptions:

the tank is adiabatic, meaning there’s no heat exchange through its walls
the tank is completely filled, lacking space for any air to cool the water
the total thermal energy in the tank is comprised of both the water’s heat energy and that of the copper block.

The mass of the remaining water in the tank can be calculated using density x volume = 1000 x 0.903 = 903 kg

The specific heat capacity for water, c = 4186 J/K-kg

Total heat content of the water Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules

The copper's mass is calculated as density x volume = 8960 x 0.097 = 869.12 kg

Copper's specific heat capacity is 385 J/K-kg

The heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules

The overall heat in the system totals up to 75.59 + 33.46 = 109.05 Mega-joules

This heat will be evenly dispersed across the system

The heat energy of the water in the system is expressed as mcT

where T signifies the final temperature

= 903 x 4186 x T = 3779958T

For copper, the heat will be

mcT = 869.12 x 385 = 334611.2T

The combined heat from both will equal the total heat of the system, meaning

3779958T + 334611.2T = 109.05 x $10^{6}$

4114569.2T = 109.05 x $10^{6}$

Thus, the final temperature T = (109.05 x $10^{6}$ )/4114569.2 = 26.5°

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