2. One hundred (100.) grams of glass beads are heated to 120.°C and then dumped

Question

2. One hundred (100.) grams of glass beads are heated to 120.°C and then dumped into the calorimeter used in the laboratory. In the calorimeter there is 150. g of water, initially at 18.0°C What is the final temperature of the water and beads? Assume that the 20.0 g stirrer is made of stainless steel with a specific heat of 0.120 cal/(gC aluminum cups with an insulator in between. Suppose that the outside one has a mass of 80.0 grams and the inner one has a mass of 40.0 grams. Look at the diagram in the laboratory manual to see which part, if any or all, of the calorimeter mass needs to be included in the calculation.

Explanation / Answer

Intial temp of galss = 120 C

water and calori meter contents = 18 C

heat lost by Glass is equal to heat gained by water + stirrer + inner cup ( calorimeter contents.)

inner outer cups are insulated and there is no heat tranfer to outside cup

Let T be the final temp

speicific heat of glass = 0.18 cal/gm-C

                     SS stirrer = 0.12

                     inner Al-cup = 0.21

                     water                =1

the heat lost by Glass = (120-T)*100*0.18

heat gained by water    = 150*1*(T-18)

                    SS stirrer    = 20*0.12*(T-18)

                    AL- inner cup = 40*0.21(T-18)

100*0.18(120-T) = 150*(T-18) +20*0.12(T-18) +40*0.21(T-18)

                    T = 28.27 0C

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