A 0.0700 kg ice cube at 30.0°C is placed in 0.517 kg of 35.0°C water in a very w

Question

A 0.0700 kg ice cube at 30.0°C is placed in 0.517 kg of 35.0°C water in a very well insulated container. The latent heat of fusion of water is 334 kJ/kg, the specific heat of ice is 2092 J/(kg · K), and the specific heat of water is 4184 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system? HINTS: This problem must be broken into pieces. Think about it chronologically: The ice can't melt until it warms up to its melting point. The ice will then melt, but will still be at zero degrees. You will need to then consider the melted ice warming to the equilibrium temperature, and the original water cooling to the equilibrium temperature.

Explanation / Answer

amount of heat absorbed by ice to change from ice at -30c to ice at 0 C
Q1 = m * Specific heat * T
For ice, specific heat = 2092 J/(kg * C)
Q1 = 0.07 * 2092 * (0-(-30) )= 4393.2 J
amount of heat required to change ice at 0 C into water at 0C,
Q 2= m * Hf,

Hf = 3.34 * 10^5 J/kg
Q = 0.07 * 3.34 * 10^5 = 23380 J

now heat requied to heat water at zero (which we got after melting of ice)to final temperature Tf is

Q3=0.07 * 4184 *
Q = 292.88 Tf

total heat absorbed= Q1+Q2+Q3=4393.2+23380+292.88Tf

Qa=27773.2+292.88Tf

now heat released from water at 35 C=Qr = m * Specific heat * T

Qr = 0.517 * 4184* (Ti – Tf)
Q r= 2163.128 * (35 – Tf) = 75709.48– 2163.128* Tf
now heat absorbed= heat released, so

27773.2+292.88Tf =75709.48– 2163.128* Tf

Tf=19.5
The final temperature is approximately 19.5C

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