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

37.8 J/k

Step-by-step 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* 10^3 J/kg\cdot K`

We know that

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

`Q=10390 J`

`S=(Q)/(T)`

Using the formula

`S=(10390)/(275)=37.8 J/K`

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

Answer 2

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

Step-by-step 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*1.039*10^(3)*(75-70)`

`Q=10390\ J`

We need to calculate the heat capacity of a sample

Using formula of heat capacity

`\Delta S=(Q)/(T)`

Put the value into the formula

`\Delta S=(10390)/(275)`

`\Delta S=37.7\ J/K`

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