A 48.6 grams copper bowl contains 53.1 g of water at 27.3 °C. A very hot copper

Question

A 48.6 grams copper bowl contains 53.1 g of water at 27.3 °C. A very hot copper cylinder, with a mass of 197.8 g, is dropped into the water, causing the water to boil, with 1.26 g being converted to steam. The final temperature of the system is 100.0°C.
Assume the specific heat capacity of water is 4.184 J/g-K, and for copper is 0.386 J/g-K, and that the latent heat of boiling of water is 2256 J/g.
How much heat was transferred to the water?


How much to the bowl ?

What is the original temperature of the cylinder?

Explanation / Answer

Given: Mass of the water = Mw = 53.1 g Mass of the water at 100oC Ms = 1.26 g Mass of the copper bowl = Mb =   48.6 g    latent heat of boiling of water is  = Lv = 2256 J/g.
Specific heat of copper bowl = Cb = 0.386 J / g - K   change of temparature = dT = 100- 27.3 = 72.7 oC specific heat of water = Cw = 4.184 J /g - K Mass of the cylinder = Mc = 197.8 g Thus, Heat transforred to water :     Qw = Cw Mw dT + Lv Ms            = (4.184 )(53.1)(72.7) + (2256)(1.26)            = 18994.34808   J    Heat transffred to the bowl Qb = Cb Mb dT         = (0.386)(48.6)(72.7)          = 1363.82292 J Thus, total heat : Q =  Qw  +  Qb                                                             = 18994.34808   J + 1363.82292 J                                  = 20358.171 J      so, Qw  +  Qb    = Mc Cb (Ti - Tf)                 20358.171 J    = (197.8)(0.386)(Ti - 100oC)                  20358.171 J = 76.3508 Ti - 7635.08                  366.639oC = Ti                   

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