Question

A piece of metal weighing 59.047 g was heated to 100.0 °C and then put it into 100.0 mL of water(initially at 23.7 °C). The metal and water were allowed to come to an equilibrium temperature,determined to be 27.8 °C. Assuming no heat lost to the environment, calculate the specific heat of themetal. (Hint: First calculate the heat absorbed by the water then use this value for “Q” to determinethe specific heat of the metal in a second calculation)

Answer 1

Specific heat of metal, c = 0.402 J/g°C

Step-by-step explanation

Given:

Mass of the metal = 59.047 g

Initial temperature of the metal = 100.0°C

Initial temperature of water = 23.7°C

Final temperature of water = 27.8°C

Final temperature of the metal = 27.8°C

Volume of water = 100.0 mL

Required: Specific heat of the metal

We know: specific heat of water = 4.184 J/g°C

Density of water = 1 g/mL

Solution

Step 1: Find the mass of water using density and volume given

Mass = density x volume

Mass = 1 g/mL x 100 mL

Mass = 100 g

Step 2: Calculate the specific heat of the metal

`Q_{metal\text{ }}=Q_(water)`

Q = m x c x Δ T

Therefore:

[m x c x Δ T] of metal = [m x c x Δ T] of water

59.047 g x c x (100-27.8 °C) = 100 g x 4.184 J/g°C x (27.8-23.7 °C)

c = 0.402 J/g°C