Question

A 2.85 g lead weight, initially at 11.0 ∘C, is submerged in 7.43 g of water at 52.2 ∘C in an insulated container. You may want to reference ( pages 349 - 355) section 10.4 while completing this problem. Part A What is the final temperature of both the weight and the water at thermal equilibrium? Express the temperature in Celsius to three significant figures. T = ∘C

Answer 1

Using, q = m c dT       -- equation 1   

Where, q = heat change                                                                 , m= mass in gram,

c = specific heat (in terms of J/g⁰C)                          

dT = final temperature – initial temperature.

Let the equilibrium temperature = T⁰C

Since, metal at low temperature is placed in water at higher temperature, the metal absorbs some heat from water. As a result, the temperature of metal increase and the temperature of water decreases.

So, change in temperature of lead = final temp – initial temp = (T – 11.0)⁰C

            change in temperature water = final temp – initial temp = (52.2 - T)⁰C

Now, heat gained by lead metal,

q1 = 2.85 g x (0.160 J/g⁰C) x (T – 11.0)⁰C

Heat lost by water,

                q2 = 7.43 g x (4.18 J/g⁰C) x (52.2 - T)⁰C

Since, heat gains by lead is equal to heat lost by water, q1 = q2

Or, 2.85 g x (0.160 J/g⁰C) x (T – 11.0)⁰C = 7.43 g x (4.18 J/g⁰C) x (52.2 - T)⁰C

Or, 0.456 J/ ⁰C x (T – 11.0)⁰C = 31.0574 J/ ⁰C x (52.2 - T)⁰C

Or, (0.456 J/ ⁰C x T) – (0.456 J/ ⁰C x 11⁰C) = (31.0574 J/ ⁰C x 52.2⁰C) - (31.0574 J/ ⁰C x T)

Or, (0.456 J/ ⁰C x T) + (31.0574 J/ ⁰C x T) = (31.0574 J/ ⁰C x 52.2⁰C) + (0.456 J/ ⁰C x 11⁰C)

Or, T x (31.5134 J/ ⁰C) = 1626.21228 J

Or, T = (1626.21228 J) / (31.5134 J/ ⁰C) = 51.604⁰C

Hence, equilibrium temperature = 51.6⁰C