Answers
In the ist part , specific heat capacity of metal Pb is not given . Without this we cannot calculate final temperature. Thus i have supposed specific heat of Pb is x. I have solved rest of the part . You have to just plug in the value of x and you will be able to calculate final equilibrium temperature.
![of Mass Molas Ag= 108 g/mol Moles of Aga moles 9.6 108 x OT X Moles Energy Requiseda Molas Heat capacity ] X 17.1 25.35 9.6 1](http://img.homeworklib.com/questions/3a885060-5f8c-11eb-8799-273b777106a6.png?x-oss-process=image/resize,w_560)
Let T Final temperature = specific Heat of Pb Metal - х Let of Ilgoc 4.18 water- Heat specific (Heat gain by Pb) loose by water) = (Heat 8.19*4.18 (53-1) - x 2.80X x (T-11) x (T-11 12.926 (53-1) = 12.226 TE 648 Ta - 11x + Il X T (x + 12.226) 648 t 648 + 11a Final Temperature 12.826 ta Final get Temperature of and x Value put the specific x Mass X KT-T;) 2 Required.
Energy Heat specific x (24.6-13.0) 45= 9.6 Heat specific Heat of Metal 0.404 319°C c
of Mass Molas Ag= 108 g/mol Moles of Aga moles 9.6 108 x OT X Moles Energy Requiseda Molas Heat capacity ] X 17.1 25.35 9.6 108 = Energy Required 38:53 ] Molas Heat capacity 4 specific Meet of Aga Molas Mass 25.35 Ilgoc lo 8 Ag specific Heat of 0.25 olgoc
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