Respuesta :
Given:
Q = 45.5 J, amount of heat absorbed
ΔT = 28.5 - 21.1 = 7.4 °C = 7.4 K, temperature change
c = 0.82 J/(g-°C), specific heat of CaCO₃.
Le m = the mass of the sample (g).
Then
Q = mcΔT
(45.5 J) = (m g)*(0.82 J/(g-°C))*(7.4 °C)
m = 45.5/(0.82*7.4) = 7.4984 g
Answer: 7.5 g (nearest tenth)
Q = 45.5 J, amount of heat absorbed
ΔT = 28.5 - 21.1 = 7.4 °C = 7.4 K, temperature change
c = 0.82 J/(g-°C), specific heat of CaCO₃.
Le m = the mass of the sample (g).
Then
Q = mcΔT
(45.5 J) = (m g)*(0.82 J/(g-°C))*(7.4 °C)
m = 45.5/(0.82*7.4) = 7.4984 g
Answer: 7.5 g (nearest tenth)
The mass of the sample of Calcium carbonate is 0.195g
The heat capacity formula,
[tex]\rm \bold{ Q= mC\Delta T}[/tex]\
Where,
Q- Heat absorbed = 45.5 J
C- specific heat capacity of [tex]\rm \bold{ CaCO_3 = 0.82 J/g/^\cdot C }[/tex]
[tex]\rm \bold{ \Delta T}[/tex]- change in temperature = 280.5 k
m - mass = ?
Solving Equation for m
[tex]\rm \bold {m = \frac{45.5 J}{0.82\times 280.5 K } }\\\rm \bold {m = 0.195 g }[/tex]
Hence , we can conclude that the mass of sample is 0.195g.
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