Question

A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 42 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?

0.171
0.342
1.717
3.420

Answer 1

1.71

Step-by-step explanation:

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Answer 2

1.717 J/g °C ( third option)

Step-by-step explanation:

A piece of the unknown metal dropped into water this means that Q of metal is equal to Q of the water. We write this equality as follows:

Step 1: writing the formulas:

Q = mc∆T

⇒ -Q(metal) = Q(water) Because :Metal dropped into water this means that Q of metal is equal to Q of the water.

We can write the formula different :

Mass of metal * (cmetal)(ΔT) = Mass of water *(cwater) (ΔT)

⇒ Here c is the specific heat and depends on material and phase

For this case :

mass of the metal = 68.6g

mass of the water = 42g

Specific heat of the metal = TO BE DETERMINED

Specific heat of the water = 4.184J/g °C

Initial temperature of the metal = 100 °C ⇒ Change of temperature: 52.1 - 100

Initial temperature of the water = 20°C ⇒ Change of temperature: 52.1 - 20

Step 2: Calculating specific heat of the metal

-(Mass of metal) * (cmetal)*(ΔT)) = (Mass of water) *(cwater)*(ΔT)

-68.6g (cmetal)(52.1 - 100) = 42g (4.184j/g °C) (52.1 - 20)

-68.6g *cmetal * (-47.9) = 42g (4.184j/g °C) *(32.1)

3285.94 * cmetal = 5640.87

cmetal = 5640.87 / 3285.94 = 1,71667 J/g °C

cMetal = 1.717 J/g °C