Question

An irregular lump of an unknown metal has a measured density of 5.19 g/mL. The metal is heated to a temperature of 179 °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 31.6 mL, and the temperature is recorded as 39.2 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Answer 1

let the density of wales be Iglml. Vol. des coader - 25 ml 25 g es wailer. Specific beat it wader - 4.186 3/900 So, heat absorbed by water is . Heat - mopoT. = 25 x .186+ (39.2 - 25) = 1486.03 J pensity B. metal - 5. 19 g/m volume & metal = water lead sise = (31.6-25) = 6.6 m Mass & metal - Density & vol. -5.19 g/m x 6.6 m) = 34.250 g Head giver oud by metal is equal to heat absorber by water. So, Heat given out by metal = 1486.03 J

Head - me BT 1486.03 - 34.250g & Cp. x (179 - 39.2) 11686.03 34. 25. 11 X (179-39.2) = 0.3103 / goc