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## Homework Statement

## Homework Equations

## The Attempt at a Solution

The question I'm currently working on:

When 0.6 kg of an object at 6 degree Celsius is put inside a beaker of water of mass 1.2Kg at 28 degree Celsius, the final temperature of water is 26 degree Celsius. Assuming no heat lost to the surroundings and the heat absorbed by the beaker is negligible, calculate the specific heat capacity of the object. (Specific heat capacity of water is 4200 J KG-1 K-1)

So far.. MY solution is:

Thermal energy, Q= specific heat capacity of water x mass of object and water x the change in temperature

which is: 4200 x 1.8 x (-8) = -60480

I am really clueless right now.. Hope you guys are able to help and guide me with this one..