Question

A metal object with mass of 24.0 g is heated to 97.0 °C and then transferred to an insulated container containing 88.5 g of water at 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 23.2 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water. specific heat:

Answer 1

Mass of water, m= 88.5 g Specificheat of water ,s = 1 cal/g.°C Change in temperature A7 = 23.2°C-20.5°C = 2.7°C Amount of heat absorbed by water, sunter = mSAT = 88.5 gx1 ca (2.7°c) 8.°C = 238.95 cal We known that, metal =-water =-238.95 cal Mass of the substance submerged, m= 24.0 g Change in temperature A7 = 23.2°C-97.0°C = -73.8°C Therefore, metal = mSAT Specific heat capacity of metal, s = Ymetal MAT -238.95 cal (24.0 g)(-73.8°C) = 0.135 cal/g.