## 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..

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mgb_phys
Homework Helper
Nearly correct,
The change in energy of the water is:
Q1 = specific heat of water * mass of WATER * change in temperature of water

Then you can use this with:
Q2 = specific heat of object * mass of OBJECT * change in temperature of object

Since no heat is lost, Q1 lost by water equals the Q2 gained by object.