Question

Determine the specific heat of a 213.0 g piece of metal in J/g•°C if the temperature of the metal drops from 70.2 °C to 25.6 °C when placed in a calorimeter with 250.0 mL of water at an initial temperature of 22.5 °C (specific heat of water is 4.184 J/g•K).

Answer 1

0.3415 J/g °C

Step-by-step explanation:

M1 = 213.0g

c1 = ?

T1 = 70.2°C

T3 = 25.6 °C

T2 = 22.5°C

C2 = 4.186 J/g°C

Vol. of water = 250mL = 250cm³

p (density of water ) = 1g/L

p = mass / volume

mass (m) = density (p) * volume (v)

m = 1 * 250 = 250g

Heat loss by metal = heat gain by water

Q1 = Q2

Mc∇T = Mc∇T

Mc (T1 - T3) = Mc* (T3 - T2)

213 * c * (70.2 - 25.6) = 250 * 4.186 * (25.6 - 22.5)

(213 * 44.6)c = 1046.5 * 3.1

9499c = 3244.15

c = 0.3415 J/g.°C

The specific heat capacity of the metal is 0.3415 J/g °C