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2 years ago

When using a different calorimeter, and mixing 50 ml of hot water at 65 degrees c with 60 ml of water in the calorimete

Physics

1 answer:

Paha777 [63]2 years ago

5 0

The specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

<h3>Conseervation of energy</h3>

The heat capacity of the calirometer is determined by applying the principle of conservation of energy.

Heat lost by the hot water = Heat gained by the calirometer

$Q _w = Q_c\\\\M_w C_w\Delta \theta _w = M_c C_c\Delta \theta _c$

where;

M is mass

mass = density x volume = ρV

Density of water = 1 g/ml

Mass of hot water = 1 x (50) = 50 g

Mass of water in calorimeter = 1 x (60) = 60 g

<h3>Equilibrium temperature</h3>

$\Delta T_c = 5.5\\\\T - 25 = 5.5\\\\T = 30.5 \ ^0C$

<h3>Specific heat capacity of the calirometer</h3>

$50 \times 4.184 \times (65 - 30.5) = 60 \times C_c \times (30.5 - 25)\\\\7217.4 = 330C_c\\\\C_c = \frac{7217.4}{330} \\\\C_c = 21.87 \ J/g^0C$

Thus, the specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

The complete question is below

When using a different calorimeter, and mixing 50 ml of hot water at

65 degrees C with 60 ml of water in the calorimeter at 25 degrees C, the temperature of the calorimeter increased by 5.5 degrees C.

a. Calculate the heat capacity of this calorimeter?

Learn more about heat capacity here: brainly.com/question/16559442

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