A 190 g copper bowl contains 130 g of water, both at 21.0degree C. A very hot 48

Question

A 190 g copper bowl contains 130 g of water, both at 21.0degree C. A very hot 480 g copper cylinder is dropped into the water, causing the water to boil, with 3.62 g being converted to steam. The final temperature of the system is 100degree C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g middot K, and of copper is 0.0923 cal/g middot K. The latent heat of vaporization of water is 539 Cal/kg. (a) Number (b) Number (c) Number Units Units

Explanation / Answer

Mass of the copper bowl = Mb = 190 g =

temparatures of the bowl and water = t = 21 deg

Mass of the water =  Mw = 130 g

Mass of the copper cylinder =  Mc = 480 g =

mass of the steam = 3.62 g = Ms

final temparature of the system is = 100 deg

A).

Heattransffred to the water :

Qw = Mw*Cw*delta(T) + Lv*Ms

= (130g)(1 cal /g)(100-21) + (539 cal /g )(3.62 g)

= 10270 cal + 1951 cal

= 12221 cal

B).

Heattransffred to bowl

Qb= Cb*Mb*delta(T)

= (0.0923 cal /gC)(190)(100-21 )

= 1385.42 cal

Let , T be the orginal temparature of the system (cylinder)

Qw + Qb = Mc*Cc*(T - 100)

T   = [ (Qw + Qb ) / Mc*Cc ] + 100

= [( 12221 +1385.42 )/ (480g )(0.0923 cal / g C )] + 100

= [13606.42 / 44.30] + 100

= 407.143 C

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