A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at $15.0 ^ { \circ } \mathrm { C }$ In a few minutes,she measures the final temperature of the system to be 40.0°C.What is the specific heat of the 400.0-g piece of metal,assuming that no significant thermal energy is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

Question

Quizplus · Accepted Answer

We can use the principle of conservation of energy in this problem. The amount of heat released by the metal is equal to the amount of heat absorbed by the water and the aluminum cup. We can write this as:

mass of metal x specific heat of metal x change in temperature of metal = (mass of water x specific heat of water + mass of aluminum cup x specific heat of aluminum cup) x change in temperature of water and aluminum cup

Substituting the given values and solving for the specific heat of the metal:

400.0 x cmetal x (100.0 - 40.0) = (100.0 x 4186.0 + 500.0 x 900.0) x (40.0 - 15.0)

cmetal = ((100.0 x 4186.0 + 500.0 x 900.0) x (40.0 - 15.0)) / (400.0 x (100.0 - 40.0))

cmetal = 2270 J/kg ∙ K

Therefore, the best choice is B (2270 J/kg ∙ K).

A Lab Assistant Drops a 400 15.0∘C15.0 ^ { \circ } \mathrm { C }15.0∘C

A Lab Assistant Drops a 400 $15.0 ^ { \circ } \mathrm { C }$