Question

1. A 450-g ball of copper at 20 0C is placed in a pot of boiling water until equilibrium is reached. How much thermal energy is absorbed by the ball?[ specific heat of copper = 385.1 J/Kg 0C]

2. A 100-g mass of an unknown material at 100 0C is placed in an aluminum calorimeter of 60.0 g that contains 150 g of water at an initial temperature of 20.0 0C. The final temperature is observed to be 21.5 0C. What is the specific heat of the substance? [specific heat of water = 4186 J/Kg 0C]

Answer 1

1. given :

m = mass copper bball = 450 g = 0.450 kg

initial temperature = Ti = 20 degree celsius

tmperature of boiling water = 100 degree celsius

specific heat of copper = s = 385.1 J/Kg /⁰C

As the boiling water taken is large surrounding, the temperature of water does not change, and thats why, the final temperature of copper will be 100 degree celsius.

Therefore, heat energy absorbed by the ball = Q = ms(Tf - Ti) = 0.450*385.1*(100-20) = 13863.6 J [answer]

2. given :

m = mass of unknow material = 100 g = 0.1 kg

m' = mass of aluminium calorimeter = 60 g = 0.06 kg

M = mass of water = 150 g = 0.15 kg

s = specific heat of unknown material = ?

s' = specific heat of aluminium = 921.1 J/kg/C

S = specific heat of water = 4186 J/Kg/C

T1 = initial temperature of unknown material = 100 degree celsius

T2 = initial temperature of water and aluminium = 20 degree celsius

T = final temperature = 21.5 degree celsius

Applying Principle of Calorimetry :

ms(T1-T) = m's'(T-T2) + MS(T-T2)

=> ms(T1-T) = (T-T2)(m's' + MS)

= 130.54 J/kg/C [answer]