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^oC

Specific heat of silver = 0.234 J/g^oC , Al = 0.910 J/g^oC , Water = 4.190 J/g^oC

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^oC*(tf-95.5^oC) = 277 g * 4.190 J/g^oC * (tf-24.5^o)+ 205 g * 0.910 J/g^oC*(tf-24.5^oC)

-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 ^oC

final temperature of the water, cup, and bear when they reach thermal equilibrium is 27.98 ^oC