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 copper pellets at a temperature of 92°C to 185 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 copper pellets and water is 25°C, how many whole pellets did the first group use in the experiment? The specific heat of copper is 0.0924 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). (A) higher (B) lower (C) the same(c) What is the equilibrium temperature of the aluminum and water mixture for the second group?
- NCSLv 76 months agoFavorite Answer
(a) heat given up by copper = heat gained by water
(m*c*ΔT)_cu = (m*c*ΔT)_h2o
n*1.5g * 0.0924kcal/kg·ºC * (92 - 25)ºC = 185g * 1.00kcal/kg·ºC * (25 - 16)ºC
n = 269 pellets
(b) Aluminum's higher specific heat means that it gives off more heat per degree change than copper. The final temperature will be (A) higher.
269 * 1.5g * 0.215kcal/kg·ºC * (92ºC - T) = 185g * 1.00kcal/kg·ºC * (T - 16ºC)
T = 40 ºC
(40.3ºC before rounding)
Note that I didn't bother to convert the masses to kg, since the gram units cancel on both sides of the equation.
Hope this helps!