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 135 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 23°C, how many whole pellets did the first group use in the experiment? The specific heat of lead is 0.0305 kcal/(kg · °C).
(b) Will the final equilibrium temperature for the second group be higher, lower, or the same as for the first group? The specific heat of aluminum is 0.215 kcal/(kg · °C).
(c) What is the equilibrium temperature of the aluminum and water mixture for the second group?

Answer 1

a)

heat lost by lead pellets = heat gain by water,

m_lead*C_lead*dT=m_water*C_water*dT

n*1.5*0.128*(92-23)=135*4.18*(23-16)

====> n=298.2

no of lead pellets, n=298

b)

the final equilibrium temperature for the second group be higher than the first group

because, the specific heat of aluminum is greater than lead.

c)

heat lost by alumimum pellets = heat gain by water,

m_al*C_Al*dT=m_water*C_water*dT

298.2*1.5*0.9*(92-T)=135*4.18*(T-16)

====> T=47.6 oC

final equilibrium temperature is T=47.6 oC