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Aleks04 [339]
3 years ago
10

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
Chemistry
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer:

heat capacity of the sample = 37.8 J/K

Explanation:

Step 1: Data given

Temperature of the sample = 275 K

The mass of liquid nitrogen = 2kg

temperature of liquid nitrogen = 70 K

The final temperature of the nitrogen is 75 K

Step 2: Calculate heat

Q = m*c*ΔT

⇒with m = the mass of liquid nitrogen = 2 kg = 2000 grams

⇒with c= the specific heat of the liquid nitrogen = 1.04 J/g*K

⇒with ΔT = the change of temperature of liquid nitrogen = T2 - T1 = 75 - 70 = 5K

Q = 2000 grams * 1.04 J/g*K * 5K

Q = 10400 J

Step 3: Calculate the heat capacity of the sample

heat capacity of the sample = 10400 J / 275 K

heat capacity of the sample = 37.8 J/K

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To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.

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