what is the answer?

Question

Substance, Specific heat (J/g·°C), Density (g/cm3)
gold	0.1256	19.3
silver	0.2386	10.5
water	4.184	1.00

A silver cube with an edge length of 2.41 cm and a gold cube with an edge length of 2.62 cm are both heated to 81.8 ∘C and placed in 111.5 mL of water at 19.8 ∘C . What is the final temperature of the water when thermal equilibrium is reached?

𝑇final=

Answer 1

For this type of problem, we'll want to use q = mC∆T

q = heat

m = mass

C = Specific heat

∆T = change in temperature

First, we'll find the mass of each cube by calculating the volume of the cube and then using the density.

Silver cube: volume = (2.41 cm)³ = 14.0 cm³. Mass of silver cube = 14.0 cm3 x 10.5 g/cm3 = 147 g

Gold cube: volume = (2.62 cm)³ = 17.98 cm³. Mass of gold cube = 17.98 cm³ x 19.3 g/cm³ = 347 g

The heat LOST by the two metals must equal the heat GAINED by the water:

q_silver + q_gold = q_H2O

(147 g)(0.2386 J/gº)(81.8 - Tf) + (347 g)(0.1256 J/gº)(81.8 - Tf) = (111.5 g)(4.184 J/gº)(Tf - 19.8)

2869 - 35.1Tf + 3565 - 43.6Tf = 466.5Tf - 9237

545.2Tf = 15671

Tf = 28.7º

(be sure to check all of the math)