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5 February, 16:02

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

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  1. 5 February, 16:21
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    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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