The specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.
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
[tex]Q _w = Q_c\\\\M_w C_w\Delta \theta _w = M_c C_c\Delta \theta _c[/tex]
where;
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
[tex]\Delta T_c = 5.5\\\\T - 25 = 5.5\\\\T = 30.5 \ ^0C[/tex]
[tex]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[/tex]
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?
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