Specific Heat Capacity

How much thermal energy is required to raise the temperature of a 2 kg aluminium block from 25 °C to 30 °C? [The specific heat capacity of aluminium is 900 Jkg-1 oC-1]

Answer:

Mass, m = 2kg
Specific heat capacity, c = 900 Jkg-1 oC-1
Temperature change, θ = 30 - 25 = 5 oC

Thermal energy required,
Q = mcθ = (2)(900)(5) = 9000J.

An electric heater supplies 5 kW of power to a tank of water. Assume all the energy supplied is converted into heat energy and the energy losses to the surrounding is negligible. How long will it take to heat 500 kg of water in the tank from 25 to 100 °C? [Specific heat capacity of water = 4200 J kg-1 oC-1]

Answer:

P = 5000W
m = 500kg
c = 4200 J kg-1 oC-1
θ = 100 - 25 = 75oC
t = ?

We assume,
all the electrical energy supplied = heat energy absorbed by the water
Pt = mcθ
(5000) t = (500)(4200)(75)
t = 31500s = 525 minutes = 8 hours 45 minutes

(Practically the time can be much longer than this because a lot of heat may be losses to the surrounding.)

A lead shot of mass 5g is placed at the bottom of a vertical cylinder that is 1m long and closed at both ends. The cylinder is inverted so that the shot falls 1 m. By how much will the temperature of the shot increase if this process is repeated 100 times? [The specific heat capacity of lead is 130Jkg-1K-1]

Answer:

m = 5g
h = 1m × 100 = 100m
g = 10 ms-2
c = 130Jkg-1K-1
θ = ?

In this case, the energy conversion is from potential energy to heat energy. We assume that all potential energy is converted into heat energy. Therefore

mgh = mcθ
gh = cθ
(10)(100) = (130) θ
θ = 7.69 oC

A 2g lead bullet is moving at 150 m/s when it strikes a wooden block and is brought to rest. Assuming all kinetic energy is converted into thermal energy and transferred to the bullet, what is the rise in temperature of the bullet as it is brought to rest? [The specific heat capacity of lead is 130 Jkg-1K-1]

Answer:

m = 2g = 0.002kg
v = 150 m/s
c = 130 Jkg-1
θ = ?

We assume all the kinetic energy is converted into heat energy

            \begin{gathered}
  \frac{1}
{2}mv^2  = mc\theta  \hfill \\
  \frac{{v^2 }}
{{2c}} = \theta  \hfill \\
  \frac{{(150)^2 }}
{{2(130)}} = \theta  \hfill \\
  \theta  = 86.54^o C \hfill \\ 
\end{gathered} 
            

What will be the final temperature if 500 cm3 of water at 0 °C is added to 200cm3 of water at 90 °C? [Density of water = 1gcm-3]

Answer:

The density of water is 1g/cm3, which means the mass of 1 cm3 of water is equal to 1g.

Let the final temperature = θ
m1 = 500g = 0.5kg
c1 = c
θ1 = θ - 0 = θ
m2 = 200g = 0.2kg
c2 = c
θ2 = 90 - θ

m1c1θ1 = m2c2θ2
(0.5) c ( θ ) = (0.2) c ( 90 - θ )
0.5 θ = 18 - 0.2 θ
0.5 θ + 0.2 θ = 18
0.7 θ = 18
θ = 25.71 oC