Question

To determine the specific heat of an object, a student heats it to 100 degrees C in boiling water. She then places the 34.5-g object in a 151-g aluminum calorimeter containing 114 g of water. The aluminum and water are initially at a temperature of 20.0 degrees C, and are thermally insulated from their surroundings. If the final temperature is 23.6 degrees C, what is the specific heat of the object?  

I have the specific heat of water as 4186 J/(kg.K)

I have the specific heat of aluminum as 900 J/(kg.K)

Please provide me with an explanation of the concept and why we use a specific formula to solve, also the actual formula, and also step by step solving and include units please all throughout the calculations.

Typed answer please, handwritten is difficult to read.

Answer 1

From the principle of calorimetry:

Heat lost by the object = Heat gained by aluminium + heat gained by water

m_object x C_object x (100^o C - 23.6^o C) = m_Alx C_Alx (23.6^o C - 20^o C) + m_wx C_wx (23.6^o C - 20^o C)

34.5 x C_object x 76.4 = 151 x 900 x 3.6 + 114 x 4186 x 3.6

solving, we get specific heat of object as:

C_object = 837.4 J/(kg.K)