Question

A 32 g piece of copper pipe at 77.9 degrees C was placed in water at 27.4 degrees C. The final temperature of the system was 28.9 degrees C. What was the volume of the water? The specific heats of copper and water are 0.385 J/g degrees C and 4.184 J/g degrees C, and the density of water is 0.998 g/mL. Assume no heat was lost to the surroundings

Answer 1

Answer:

Explanation:

Heat lost by hot copper = heat gained by water= mass x specific heat x change in temperature.

Heat lost by piece of copper

= 32 x 0.385 x ( 77.9 - 28.9)

=603.68 J

If m be the mass of water taken

Heat gained by water

=m x 4.184 x (28.9 - 27.4 )

= 6.276 m

Heat lost = heat gained

6.276 m = 603.68

m = $\frac{603.68}{6.276}$

= 96.188 g

Volume of water = mass / density

= $\frac{96.188}{0.998}$

= 96.38 mL