Question

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

Answer 1

Answer:

The metal has a specific heat of 0.34333 J / g °C

Explanation:

The 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