Question

A tissue sample at 275K 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.)

Answer 1

Assume no heat is lost to the surrounding air.

We know that, Q = m_N C_N (T_f - T_i)

where, m_N = mass of liquid nitrogen = 2 kg

C_N = specific heat of liquid nitrogen = 1.039 kJ/kg.K

T_f = final temperature of nitrogen = 75 K

T_i = initial temperature of nitrogen = 70 K

then, we get

Q = (2 kg) (1.039 x 10³ J/kg.K) [(75 K) - (70 K)]

Q = [(2078 J/K) (5 K)]

Q = 10390 J

The heat capacity of a sample which will be given by -

S = Q / T [(10390 J) / (275 K)]

S = 37.7 J/K