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Sagot :
We have that the energy (in calories) is transferred to the water as heat,to the bowl and the original temperature of the cylinder is mathematically given as
- Qw=20.3 kcal
- Q= 1.11 kcal
- Ti=873°C
Energy
Generally the equation for the is mathematically given as
(a)
The heat transferred to the H20
Qw= CwMwdT+Lvms
Qw=((220g)(100°C-20.0T)+(539 caVg)(5.00 g)
Qw=20.3 kcal .
(b)
The heat transferred to the bowl is
Qb= CbmbdT
Q= (0.0923 cal/gC)(150g)(100°C-20.0°C)
Q= 1.11 kcal
(c)
original temperature of the cylinder
-Qw- Qb = CcMc(T2-T1)
[tex]T1=\frac{Qw+Qm}{CcMc}+T2[/tex]
T1=873C
For more information on energy visit
https://brainly.com/question/21222010
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