Figure 1 shows a perfectly insulated cylinder containing 0.050 kg of liquid nitrogen at a temperature of 70 K - AQA - A-Level Physics - Question 1 - 2019 - Paper 2

To find the specific heat capacity, we first need to calculate the total energy supplied to the nitrogen using the formula:

$E = P imes t$

where:

• $P = 12 \, W$ (power of the heater)
• $t = 890 \, s$ (time)

Substituting the values, we get:

$E = 12 \, W imes 890 \, s = 10680 \, J$

Next, we must calculate the energy required to convert the nitrogen from a liquid to a gas:

1. Determine the mass of nitrogen: $m = 0.050 \, kg$

2. The energy required to vaporize the nitrogen is given by:

   $Q_{vap} = m imes L_v$

   where:

   • $L_v = 2.0 imes 10^5 \, J \, kg^{-1}$ (specific latent heat of vaporization)

   So,

   $Q_{vap} = 0.050 \, kg imes 2.0 imes 10^5 \, J \, kg^{-1} = 10000 \, J$

Now we can find the specific heat capacity $ext{c}$ using:

$Q_{total} = Q_{vap} + Q_{heating}$

Where:

• $Q_{heating} = m imes c imes ext{temperature change}$

The temperature change from 70 K to 77 K is:

$ext{temperature change} = 77 \, K - 70 \, K = 7 \, K$

Thus,

$Q_{heating} = 0.050 \, kg imes c imes 7 \, K$

Now we have,

$10680 \, J = 10000 \, J + 0.050 \, kg imes c imes 7 \, K$

Rearranging to find $ext{c}$:

$680 \, J = 0.050 \, kg imes c imes 7 \, K$

c = rac{680 \, J}{0.050 \, kg imes 7 \, K} = 1944.29 \, J \, kg^{-1} \, K^{-1}

Thus, the specific heat capacity of liquid nitrogen is approximately $1944.29 \, J \, kg^{-1} \, K^{-1}$.