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A 150g copper bowl contains 220g of water, both at 20.0oC, A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100oC, Neglect energy transfers with the environment.
a) How much energy (in calories) is transfered to the water as heat?
b) How much to the bowl?
c) What is the original temperature of the cylinder?

Answer :

OKPALAWALTER8IN

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

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