Answer:
1.717 J/g °C ( third option)
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
Answer:
1.71
Explanation:
GOT IT RIGHT ON ODDESYWARE ;) Have a nice day
