Question

1. A 2.80 g lead weight, initially at 11.0 ∘C, is submerged in 8.19 g of water at 53.0 ∘C in an insulated container.

What is the final temperature of both the weight and the water at thermal equilibrium?

2. It takes 45.0 J to raise the temperature of an 9.60 g piece of unknown metal from 13.0∘C to 24.6 ∘C. What is the specific heat for the metal?

3. The molar heat capacity of silver is 25.35 J/mol. ∘C. How much energy would it take to raise the temperature of 9.60 g of silver by 17.1 ∘C?

4. What is the specific heat of silver?

Answer 1

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.
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