Question

A 72.0-gram piece of metal at 96.0 °C is placed in 130.0 g of water in a calorimeter at 25.5 °C. The final temperature in the calorimeter is 31.0 °C. Determine the specific heat of the metal. Show your work by listing various steps, and explain how the law of conservation of energy applies to this situation.

Answer 1

Answer: The answer is S = 0.1528 cal/g °C

Explanation:

By the law of conservation of energy, energy is neither created nor destroyed.

So, energy lost by metal pieces is equal to the energy gained by water in the calorimeter.

Specific heat of water is 1 cal/g °C

⇒ heat energy  Q = mSΔT, where m = mass of a substance

                                                        S = specific heat

                                                        ΔT = change in temperature

Now, the heat lost by metal piece, Q = 72×S×(96-31)

                                                      = 4680×S cal

Heat gained by water, Q = 130×1×(31-25.5)

                                        = 715 cal

⇒ 4680×S = 715.

⇒ S = 0.1528 cal/g °C.