Question

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitrogen is 75 K. What is the heat capacity of the sample in J/K? (Note: Assume no heat is lost to the surrounding air.) A.0.05 J/K

Answer 1

Answer:

The heat capacity of a sample is 37.7 J/K.

Explanation:

Given that,

Submerged temperature of tissue sample = 275 K

Mass of liquid nitrogen= 2 kg

Temperature = 70 K

Final temperature = 75 K

We need to calculate the heat

Using formula of heat

$Q=mc(T_{f}-T_{i})$

Put the value into the formula

$Q=2\times1.039\times10^{3}\times(75-70)$

$Q=10390\ J$

We need to calculate the heat capacity of a sample

Using formula of heat capacity

$\Delta S=\dfrac{Q}{T}$

Put the value into the formula

$\Delta S=\dfrac{10390}{275}$

$\Delta S=37.7\ J/K$

Hence, The heat capacity of a sample is 37.7 J/K.

Answer 2

Answer:

37.8 J/k

Explanation:

Temperature =T=275 K

Mass of liquid nitrogen=2 kg

Initial temperature=$T_1=70 K$

Final temperature=$T_2=75 K$

We have to find the heat capacity of the sample in J/K.

Specific heat of liquid nitrogen=$c=1.039\times 10^3 J/kg\cdot K$

We know that

$Q=mc\Delta T=2\times 1.039\times 10^3\times (75-70)$

$Q=10390 J$

$S=\frac{Q}{T}$

Using the formula

$S=\frac{10390}{275}=37.8 J/K$

Hence, the heat capacity of sample=37.8 J/k