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Soloha48 [4]
3 years ago
6

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

ogen 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
Physics
2 answers:
kirill115 [55]3 years ago
7 0

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.

Romashka-Z-Leto [24]3 years ago
5 0

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

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