Question

A 401−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assuming no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?

Answer 1

m(water) = 159.0 g

T(water) = 22.8 oC

C(water) = 4.184 J/goC

m(copper) = 401.0 g

T(copper) = 89.5 oC

C(copper) = 0.3878 J/goC

T = to be calculated

We will be using heat conservation equation

Let the final temperature be T oC

use:

heat lost by copper = heat gained by water and 3

m(copper)*C(copper)*(T(copper)-T) = m(water)*C(water)*(T-T(water)) + C3*(T-T(water))

401.0*0.3878*(89.5-T) = 159.0*4.184*(T-22.8)+10.0*(T-22.8)

155.5078*(89.5-T) = 675.256*(T-22.8)

13917.9481 - 155.5078*T = 675.256*T - 15395.8368

T= 35.3 oC

Answer: 35.3 oC