Question

You heat a 5.2 gram lead ball (specific heat capacity = 0.128 Joules/g-deg) to 183. ∘ C. You then drop the ball into 34.5 milliliters of water (density 1 g/ml; specific heat capacity = 4.184 J/g-deg ) at 22.4 ∘ C. What is the final temperature of the water when the lead and water reach thermal equilibrium?

Answer 1

Answer:

The final temperature when the lead and water reach thermal equilibrium is $23.14^{o} C$

Explanation:

Using the law of conservation of energy the heat lost by the lead ball is gained by the water.

$-q_{L} =+q_{w}$ .............................1

where

$q_{L}$ is the heat energy of the lead ball;

$q_{w}$ is the heat energy of water;

but q = msΔ

T...............................2

substituting expression in equation 2 into the equation 1;

Δ

T =Δ

T .............................3

where $m_{L}$ is the mass of the lead ball = 5.2 g

$s_{L}$ is the specific heat capacity of the lead ball = 0.128 Joules/g-deg

$m_{w}$ is the mass of water = volume x density = 34.5 ml x 1 g/ml = 34.5 g

$s_{w}$ is the specific heat capacity of water = 4.184 J/g-deg

Δ

T is the temperature changes encountered by the lead ball and water   respectively

inputting the values of the parameters into equation 3

5.2 g x 0.128 Joules/g-deg x (183-22.4) = 34.5 g x 4.184 J/g-deg x ($T_{2}$ -22.4)

$T_{2}$ -22.4 = $\frac{106.9}{144.3}$

$T_{2}$ -22.4 = 0.7408

$T_{2}$ = $23.14^{o} C$

Therefore the final temperature  when the lead and water reach thermal equilibrium is $23.14^{o} C$