Since LF,ice = 333 J/g, what is the change in entropy (in cal/K) when 1.00 kg of ice melts?
A)144
B)291
C)312
D)516
E)80

Question

Quizplus · Accepted Answer

To calculate the change in entropy (ΔS) during the melting of 1.00 kg of ice, we can use the equation:
ΔS = q/T
where q is the amount of heat absorbed during the melting process and T is the temperature at which the process occurs.

The amount of heat absorbed is given by:
q = m × LF
where m is the mass of the ice and LF is the latent heat of fusion. Thus, q = 1.00 kg × 333 J/g = 333000 J.

To convert the units of ΔS from J/K to cal/K, we divide by 4.184:
ΔS = (333000 J/K) / (4.184 J/cal·K) = 79557 cal/K

Therefore, the answer is B) 291.

Since LF,ice = 333 J/g, What Is the Change in Entropy