In a physics lab students are conducting an experiment to learn about the heat c

Question

In a physics lab students are conducting an experiment to learn about the heat capacity of different materials. The first group is instructed to add 1.5-g lead pellets at a temperature of 92°C to 145 g of water at 16°C. A second group is given the same number of 1.5-g pellets as the first group, but these are now aluminum pellets. Assume that no heat is lost to or gained from the surroundings for either group.

(a) If the final equilibrium temperature of the lead pellets and water is 24°C, how many whole pellets did the first group use in the experiment? The specific heat of lead is 0.0305 kcal/(kg · °C). : ? pellets

(b)What is the equilibrium temperature of the aluminum and water mixture for the second group?: ? °C

Explanation / Answer

Hear lost by lead pellets= heat gained by water

mass of lead pellets* specific heat* temperature difference =   mass of water* specific heat* temperature difference

n*1.5*0.0305 *(92-24)= 145*1*(24-16) ( specific heat of water= 1 Ca/g.deg.c) n= number of pelltets

n= 145*1*8/ (1.5*0.0305*68) =373 pellets

b) specific heat of aluminium =0.215 Cal/g.deg.c

1.5*0.215*(92-T)= 145*1*(T-16)

0.3225*(92-T)= 145T- 2320

T*(145+0.3225)= 2320+0.3225*92

T*145.3225= 2320+29.3=2349.3

T= 2349.3/145.3225, T = 16.6 deg.c

b)

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