In a coffee-cup calorimeter, 50.0 g hot water at 60.0 C was mixed with 50.0 g cold water at 20.0 C. If the final temperature is 36.3, what is the heat capacity of the calorimeter in J/C? Specific heat of water is 4.184 J/(g C)

Respuesta :

Answer:

2.88 J/g°C

Explanation:

Mass calorimeter (Mc) = 50g

Temperature of calorimeter (T1) = 60°C

Mass of water (Mw) = 50g

Specific heat capacity of water (Cw) = 4.184 J/g°C

Specific heat capacity of calorimeter (Cc) = ?

Temperature of water (T2) = 20°C

Final temperature of the content (T3) = 36.6°C

Assuming no heat lose from the set up

Heat loss by copper calorimeter = heat gain by the water

Mc * Cc * (T1 - T3) = Mw * Cw * (T3 - T2)

50 * Cc * (60 - 36.3) = 50 * 4.184 * (36.3 - 20)

50 * Cc * 23.7 = 209.2 * 16.3

1185Cc = 3409.96

Cc = 3409.96 / 1185

Cc = 2.877 J/g°C

Cc = 2.88 J/g°C

The specific heat capacity of the copper calorimeter is 2.88 J/g°C

The heat capacity of the calorimeter will be "2.88 J/g°C".

Heat and Temperature

According to the question,

Mass calorimeter, [tex]M_c[/tex] = 50 g

Calorimeter's temperature, [tex]T_1[/tex] = 60°C

Water's mass, [tex]M_w[/tex] = 50 g

Water's specific heat capacity, [tex]C_w[/tex] = 4.184 J/g°C

Water's temperature, T₂ = 20°C

Content's final temperature, T₃ = 36.3°C

We know,

Heat loss by copper calorimeter = Heat gain by the water

→ [tex]M_c[/tex] × [tex]C_c[/tex] × (T₁ - T₃) = [tex]M_w[/tex] × [tex]C_w[/tex] × (T₃ - T₂)

By substituting the values, we get

  50 × [tex]C_c[/tex] × (60 - 36.3) = 50 × 4.184 × (36.3 - 20)

            50 × [tex]C_c[/tex] × 23.7 = 209.2 × 16.3

                                [tex]C_c[/tex] = [tex]\frac{3409.96}{1185}[/tex]

                                     = 2.88 J/g°C

Thus the response above is correct.

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