Question

nment Score: 1438/2000 Resources Ex Give Up? Feedback Resume estion 3 of 20 > Attempt 1 - An irregular lump of an unknown metal has a measured density of 2.50 g/mL. The metal is heated to a temperature of 153 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 33.2 mL, and the temperature is recorded as 37.9 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings. 1.75 g. °C careers about us privacy policy terms of use contact us help OPONOVO XOGC

Answer 1

Ans :

Volume of lump = final volume - initial volume

= 33.2 mL - 25.0 mL

= 8.2 mL

density = mass / volume

2.50 g/mL = m / 8.2 mL

mass = 2.50 g/mL x 8.2 mL

= 20.5 g

heat lost by lump will be equal to heat gained by water

heat = mass x specific heat x change in temperature

specific heat of water = 4.184 J/g^oC

density of water is 1.00 g/mL , so mass of 25.0 mL water = 25.0 g

putting values :

20.5 g x c x (153^oC - 37.9^oC ) = 25.0 g x 4.184 J/g^oC x ( 37.9^oC - 25.0^oC)

2359.55 g.^oC x c = 1349.34 J

c = 0.5719 J/g^oC