Answer :
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
[tex]Q=mc(T_{f}-T_{i})[/tex]
Put the value into the formula
[tex]Q=2\times1.039\times10^{3}\times(75-70)[/tex]
[tex]Q=10390\ J[/tex]
We need to calculate the heat capacity of a sample
Using formula of heat capacity
[tex]\Delta S=\dfrac{Q}{T}[/tex]
Put the value into the formula
[tex]\Delta S=\dfrac{10390}{275}[/tex]
[tex]\Delta S=37.7\ J/K[/tex]
Hence, The heat capacity of a sample is 37.7 J/K.
Answer:
37.8 J/k
Explanation:
Temperature =T=275 K
Mass of liquid nitrogen=2 kg
Initial temperature=[tex]T_1=70 K[/tex]
Final temperature=[tex]T_2=75 K[/tex]
We have to find the heat capacity of the sample in J/K.
Specific heat of liquid nitrogen=[tex]c=1.039\times 10^3 J/kg\cdot K[/tex]
We know that
[tex]Q=mc\Delta T=2\times 1.039\times 10^3\times (75-70)[/tex]
[tex]Q=10390 J[/tex]
[tex]S=\frac{Q}{T}[/tex]
Using the formula
[tex]S=\frac{10390}{275}=37.8 J/K[/tex]
Hence, the heat capacity of sample=37.8 J/k