Question

A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 154-g cube of copper that is initially at 84°C, and the other is a chunk of aluminum that is initially at 5.0°C. To the surprise of the student, the water reaches a final temperature of 25°C, precisely where it started. What is the mass of the aluminum chunk?

Answer 1

In this process, by principle of calorimetry, the steel container and water would not gain or loss any heat as their initial and final temperatures are same, that is, 25o C, but the copper cube with fall in temperature would loss heat and the aluminium chunk with rise in temperature would gain heat, so, at thermal equilibrium,

Heat lost by copper cube = Heat gain by aluminium chunk

Or, mcu*ccu*(84-25) = mal*cal*(25-5)

Or, 154*0.385*59 = mal*0.902*20

Or, 3498.11 = mal*18.04

Thus, mass of chunk, mal = (3498.11/18.04) gm

= 193.91 gm

Here, ccu and cal are specific heat of copper cube and aluminium chunk respectively in J/g.oC