Question

Just really confused with PART C!! I have already solved part A and B. Please help me with C I am super confused! Thanks

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 215 g of water at 16°C. A second group is given the same number of 1.5-9 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 27°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) 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). higher O lower O the same (c) What is the equilibrium temperature of the aluminum and water mixture for the second group? °C

Answer 1

PART C:

From part A, we have, no.of pellets used = 795 pellets

Hence, total mass of aluminium, m_Al = 795 * 1.5 g = 1192.9 g = 1.1929 kg

Mass of water, m_w = 0.215 kg

Heat capacity of water, C_w = 1 kcal/kg^oC

We know that, heat lost by aluminium = heat gained by water.

Let "T" be the equilibrium temperature:

Q_water = Q_Al

m_wC_w(T - T₁) = m_AlC_Al(T₂ - T)

(0.215*1) * (T - 16) = (1.1929*0.215) * (92 - T)

Solving for T, we get

T = 57.34^oC ----------- (**Answer**)

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