Question

You drop a 297-g silver figure of a polar bear into the 205-g aluminum cup of a well-insulated calorimeter containing 277 g of liquid water at 24.5°C. The bear\'s initial temperature is 95.5°C. What is the final temperature of the water, cup, and bear when they reach thermal equilibrium? The specific heats of silver, aluminum, and liquid water are, respectively, 234 J/(kg·K), 910 J/(kg·K), and 4190 J/(kg·K).

Answer #1

**Answer** – We are given, mass of silver = 297 g ,
mass of Al cup = 205 g, mass of water = 277 g , ti = 24.5
^{o} C , initial temp bears = 95.5^{o}C

Specific heat of silver = 0.234 J/g^{o}C , Al = 0.910
J/g^{o}C , Water = 4.190 J/g^{o}C

We know,

Heat loss of silver figure of a polar bear = Heat gain by water + Al

For silver figure of a polar bear, q= - m1*C_{Ag}*∆t

For water , q = m2*C_{water}*∆t

For aluminum, q = m3*C_{Al}*∆t

We know,

- m1*C_{Ag}*∆t = m2*C_{water}*∆t +
m3*C_{Al}*∆t

-297 g * 0.234 J/g^{o}C*(tf-95.5^{o}C) = 277 g *
4.190 J/g^{o}C * (tf-24.5^{o})+ 205 g * 0.910
J/g^{o}C*(tf-24.5^{o}C)

-69.498 tf + 6637.06 = 1160.63tf-28435.44 + 186.55tf - 4570.5

-69.498 tf – 1160.6 tf – 186.55 tf = -6637.06 -28435.44 - 4570.5

-1416.6 tf = -39643

So, tf = 27.98 ^{o}C

final temperature of the water, cup, and bear when they reach
thermal equilibrium is **27.98 ^{o}C**

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