A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 8.4 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?
Answers
Step-by-step explanation:
192×S×(100−21.5)
=128×394×(21.5−8.4)
+240×4200×(21.5−8.4)
⟹ S = 916
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Answer:
The metal has a specific heat of 0.34333 J / g °C
Step-by-step explanation:
he amount of heat lost by the metal is equal to the amount heat won by the water
q(lost, metal) = q(gained, water)
Step 1: Given data
q = m*ΔT *Cp
⇒with m = mass of the substance
⇒with ΔT = change in temp = final temperature T2 - initial temperature T1
⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)
mass of the metal = 68.6g
mass of the water = 8.4g
initial temperature metal = 100°C
final temperature metal = 52.1°C
initial temperature water = 20°C
final temperature water = 52.1 °C
Specific heat of water = 4.184 J/g °C
Step 2:
For this situation : we get for q = m*ΔT *Cp
q(lost, metal) = q(gained, water)
- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)
-68.6 * (52.1 - 100) * Cpmetal = 8.4g * (52.1 - 20) * 4.184 J /g °C
-68.6 * (-47.9) * Cpmetal = 1128.17
Cpmetal = 1128.17 / (-68.6 *-47.9) = 0.34333 J / g °C
The metal has a specific heat of 0.34333 J / g °C