Statistical Thermodynamics: Zeroth law of thermodynamics

Classical Mechanics Level 1

Calculate the temperature increase when $400 \text{ J}$ of heat is applied to $46.8 \text{ g}$ of $\ce{NaCl}.$

Details and assumptions:

The molar heat capacity for $\ce{NaCl}$ is $C_p = 50 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} .$
The formula weight of $\ce{NaCl}$ is $58.5 \text{ g/mol}.$

10 \text{ K}

20 \text{ K}

30 \text{ K}

40 \text{ K}

1 solution

Galactic Talk
Apr 2, 2021

here, we have to find the change in temp. when 400J of heat is applied we know, heat = change in temp x Specific heat capacity Q= del(T) x C

we have given Q = 400 and Cp we can calculate

 C = Cp x m/M
 C= 50 x 46.8/58.5
 C = 40

m= given mass M= molar mass

so change in temp. del(T)= $\frac{Q}{C}$ del(T)= $\frac{400}{40}$ del (T)= 10K