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The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

What will be the equilibrium temperature when a 235 g block of copper at 255 ∘C is placed in a 135 g aluminum calorimeter cup containing 825 g of water at 16.0 ∘C? Express your answer using three significant figures.

Answer :

samuelonum1

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

Lanuel

By applying the law of conservation of heat energy, the equilibrium temperature is equal to 21.97°C.

Given the following data:

  • Specific heat capacity of copper = 390 J/kg°C
  • Specific heat capacity of water = 4186 J/kg°C
  • Specific heat capacity of aluminum = 900 J/kg°C
  • Mass of copper = 235 grams
  • Mass of aluminum = 135 grams
  • Mass of water = 825 grams
  • Temperature of copper = 255°C
  • Temperature of water and aluminum calorimeter cup = 16.0°C

To determine the equilibrium temperature, we would apply the law of conservation of heat energy:

The quantity of heat energy lost by copper = The quantity of heat energy gained by water and aluminum calorimeter cup.

[tex]Q_{lost} = Q_{gained}[/tex]

Mathematically, this is given by the formula:

[tex]M_cC_c(T_c - T_e)=(M_wC_w + M_aC_a)(T_e-T_{wa})[/tex]

Substituting the given parameters into the formula, we have;

[tex]0.235 \times 390\times(255-T_e)=([0.825\times4186]+[0.135\times900])(T_e -16)\\\\91.65(255-T_e)=(3453.45+121.5)(T_e -16)\\\\23370.75-91.65T_e=3574.95(T_e -16)\\\\23370.75-91.65T_e=3574.95T_e -57199.2\\\\3574.95T_e+91.65T_e=23370.75+57199.2\\\\3666.6T_e=80569.95\\\\T_e=\frac{80569.95}{3666.6} \\\\T_e=21.97[/tex]

Equilibrium temperature = 21.97°C

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