In a physics lab students are conducting an experiment to team about the heat ca

Question

In a physics lab students are conducting an experiment to team about the heat capacity of different materials. The first group is instructed to add 1.5-g copper pellets at a temperature of 92degreeC to 295 g of water at 16degreeC. 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 gamed from the surroundings for either group. If the final equilibrium temperature of the copper pedets and water is 26degreeC, how many whole pellets did the first group use in the experiment? The specific heat of copper is 0.0924 kcat/(kg degreeC). No heat is exchanged with the surroundings by the system of water and pellets. How is the quantity of heat gamed by the cold water related to the quantity of heat lost by the hot pellets' pellets 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 degreeC). What is the equilibrium temperature of the alum mum and water mixture for the second group?

Explanation / Answer

Weight of coppoer pallets Wc = n * 1.5g where n is the number of pallets

Temperature of copper pallets Tc=92°C

Weight of water Ww=295g

Temperaure of water Tw=16°C

Equillibrium temperature Te = 26°C

specific heat of copper cc = 0.0924 kcal/(kgoC)

specific heat of water cw = 1 kcal/(kgoC)

at equillibrium after mixing copper in water

cc Wc (Tc-Te) = cw Ww (Te-Tw)

0.0924 * n * 1.5 * (92-26) = 1 * 295 * (26-16)

solving for n we get

n = 322.49

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