Question

Let A be the last two digits, let B be the last digit, and let C be the sum of the last three digits of your 8-digit student ID. Example: for 20245347, A = 47, B = 7, and C = 14.

A container with (15.0 + A) g of water at (8.0 + C) ^oC is placed in a freezer. How much heat must be removed from the water to turn it to ice at –(5.0 + B) ^oC? Ignore the heat capacity of the container. Give your answer in kilo-joules (kJ) with 3 significant figures.

Specific heat of ice: 2.090 J/g K

Specific heat of water: 4.186 J/g K

Latent heat of fusion for water: 333 J/g

Answer 1

Note - I will be using given student's ID for this solution..............

Note - I will be using given student's ID for this solution..............

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m = 15 + A = 15 + 47 = 62 g

T_i = 8 + C = 22 ^oC

First, we need to take this water from 22 ^oC to 0 ^oC

so,

Q = mcT

Q = 62 * 4.186 * (0 - 22)

Q = 5709.704 J

and

then we need to convert this water at 0 ^oC to ice at 0 ^oC

Q = mL

Q = 62 * 333

Q = 20646 J

and

now energy required to go from 0 ^oC to -12 ^oC

so,

Q = 62 * 2.090 * 12

Q = 1554.96 J

so

total energy to be remove

Q = 27910.66 J

Q = 27911 J