A <10.0+A> g ice cube at -15.0oC is placed in <125+B> g of water at 48.0oC. Find

Question

A <10.0+A> g ice cube at -15.0oC is placed in <125+B> g of water at 48.0oC. Find the final temperature of the system when equilibrium is reached. Ignore the heat capacity of the container and assume this is in a calorimeter, i.e. the system is thermally insulated from the surroundings. Give your answer in oC 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

A=0 B=7

Please show how to get the answer so I understand how to tackle this problem

Explanation / Answer

Q(lost) = Q(gain)

Mw Cw (Tw - Tf) = Mi Ci (Tf - Ti) + Mi Hf

0.125 x 4186 x (48 - Tf) = 0.010 x 2090 (Tf + 15) + 0.01 x 333000

25116 - 523.25 Tf = 20.9 Tf + 313.5 + 3330

Tf (544.15) = 25116 - 3643.5 = 21472.5

Tf = 39.46 deg C

Hence, Tf = 39.46 deg C

[solved using SI units of quantities given, mass in kg specific heats in J/kg-deg C for simplicity]

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