An 80-g aluminum calorimeter contains 380 g of water at an equilibrium temperature of 20°C. A 120-g piece of metal, initially at 352°C, is added to the calorimeter. The final temperature at equilibrium is 32°C. Assume there is no external heat exchange. The specific heats of aluminum and water are 910 J/kg·K and 4190 J/kg·K, respectively. The specific heat of the metal is closest to

Question

Quizplus · Accepted Answer

First, we need to use the principle of conservation of energy:
The heat lost by the hot metal = the heat gained by the aluminum calorimeter and the water
This can be represented as:
(metal mass)(metal specific heat)(final temperature - initial temperature) = (aluminum calorimeter mass)(aluminum specific heat)(final temperature - initial temperature) + (water mass)(water specific heat)(final temperature - initial temperature)
Plugging in the given values and solving for the specific heat of the metal:
(120 g)(C)(352°C - 32°C) = (80 g)(910 J/kg·K)(32°C - 20°C) + (380 g)(4190 J/kg·K)(32°C - 20°C)
Where C represents the specific heat of the metal.
Simplifying and solving for C gives:
C ≈ 520 J/kg·K
Therefore, the closest option is A) 520 J/kg·K.

An 80-G Aluminum Calorimeter Contains 380 G of Water at an Equilibrium