Question

A 248 g piece of copper initially at 314 c is dropped into 390 mL of water initially at 22.6 C Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.

​show every step please, thank you.

Answer 1

We use the principle of calorimetry to solve the problem:
It states that if no heat is lost to the surroundings, heat lost by hot body = heat gained by cold body.

Also, Heat gained by a body = mcdT,

where m is the mass,

c is the specific heat of the body,

dT = temperature change

Let the final temperature of the experiment be x

Specific heat of Cu = 0.385 J / g⁰C

Specific heat of Water = 4.186 J / g⁰C

Therfore, equating the symbols with their values from the question:
m₁c₁dT₁ = m₂c₂dT₂

=> 248 g * 0.385 J / g⁰C * ( 314 - x ) = 390 ml * 1 g / ml * 4.2 J / g⁰C ( x -22.6)

=> 29980.72 - 95.48x = 1638x - 37018.8

=> 1733.48x = 66999.52

=> x = 66999.52 / 1733.48 = 38.65 ⁰C

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