Respuesta :
Answer:
[tex]396.65[/tex] JC⁻¹
Explanation:
[tex]m_{a}[/tex] = mass of water added to calorimeter = 94.8 g
[tex]T_{ai}[/tex] = initial temperature of the water added = 60.4 C
[tex]c_{w}[/tex] = specific heat of water = 4.184 Jg⁻¹C⁻¹
[tex]m_{c}[/tex] = mass of water available to calorimeter = 94.8 g
[tex]T_{ci}[/tex] = initial temperature of the water in calorimeter = 22.3 C
[tex]T_{f}[/tex] = final equilibrium temperature = 35 C
[tex]Q_{C}[/tex] = Heat gained by calorimeter
Using conservation of heat
Heat gained by calorimeter = Heat lost by water added - heat gained by water in calorimeter
[tex]Q_{C} = m_{a} c_{w} (T_{ai} - T_{f}) - m_{c} c_{w} (T_{f} - T_{ci})[/tex]
[tex]Q_{C} = (94.8) (4.184) (60.4 - 35) - (94.8) (4.184) (35 - 22.3)[/tex]
[tex]Q_{C} = 5037.4[/tex] J
[tex]T[/tex] = Change in temperature of calorimeter
Change in temperature of calorimeter is given as
[tex]T = T_{f} - T_{ci}[/tex]
[tex]T = 35 - 22.3[/tex]
[tex]T = 12.7[/tex] C
Heat capacity of calorimeter is given as
[tex]c_{cm} = \frac{Q_{C}}{T}[/tex]
[tex]c_{cm} = \frac{5037.4}{12.7}[/tex]
[tex]c_{cm} = 396.65[/tex] JC⁻¹
The heat capacity of the calorimeter depends on the heat gained by it which is determined by the law of heat conservation.
The heat capacity of the calorimeter is 396.64 J/°C.
What is the heat capacity?
The heat capacity of an object is defined as the number of heat units needed to raise the temperature of the object by one degree.
Given:
- For water available:
The mass m1 is 94.8 g and the initial temperature t1 is 22.3 degrees.
- For water added:
The mass m2 is 94.8 g and the initial temperature t2 is 60.4 degrees.
The final equilibrium temperature t_f is 35 degrees and the specific heat of the water h= 4.184 J/g°C.
Here the law of conservation of heat is applied. According to this law, the heat gained by the calorimeter is equivalent to the difference of heat lost by the added water and heat gained by the water available in the calorimeter.
[tex]H = m_2h(t_2 - t_f) - m_1h(t_f-t_1)[/tex]
Here H is the heat gained by the calorimeter.
[tex]H = 94.8\times 4.184 ( 60.4-35) - 94.8\times 4.184 (35 - 22.3)[/tex]
[tex]H = 10074.74 - 5037.36[/tex]
[tex]H = 5037.38\;\rm J[/tex]
The change in the temperature of the calorimeter is,
[tex]T = t_f - t_1[/tex]
[tex]T = 35-22.3[/tex]
[tex]T = 12.7^\circ[/tex]
The heat capacity of the calorimeter is given as,
Heat Capacity = [tex]\dfrac {H}{T}[/tex]
Heat Capacity = [tex]\dfrac {5037.38 }{12.7}[/tex]
Heat Capacity = 396.64 J/°C.
Hence we can conclude that the heat capacity of the calorimeter is 396.64 J/°C.
To know more about the heat capacity, follow the link given below.
https://brainly.com/question/1105305.
