Question

A 5.00g piece of metal is heated to 100.0°C, then placed in a beaker containing 20.0 of water at 10.0°C. The temperature of the water rises to 15.0°C. Calculate the specific heat of the metal.

Answer 1

Answer:

This metal has a specific heat of 0.9845J/ g °C

Explanation:
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)

Step 2: Calculate specific heat
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)
-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =
-5 * (-85)(Cpmetal) = 418.4

Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C