6 A 470g ofan unknown metal was heated to 1000C and the hot

(6) A 470-g ofan unknown metal was heated to 100.0°C and the hot metal was immediately added into 30.0 g of cold water in a Styrofoam calorimeter. The initial temperature of cold water and calorimeter was 22.0 After metal, water and calorimeter have come to thermal equilibrium, their final temperature was 32.2 (a) Calculate the amount of heat (in Joules) absorbed by water? (b) Calculate the amount of heat absorbed by calorimeter, which has heat capacity of 25 J/°C (c) What is the total amount of heat lost by hot metal? (d) Calculate the specific heat of metal

Solution

specific heat of water= 4.184 J/gm.deg.c

heat absorbed by water= mass of water* specific heat of water* change in temperature

= 30gm* 4.184j/gm.deg.c*(32.2-22)= 1280.3 joules

heat absorbed by calorimeter= heat capacity of calorimeter* temperature difference= 25*(32.2-22) joules =305 joules

for adiabatic system, this much heat has to be lost by metal, heat lost by metal= -(1280.3+305)=-1585.3 joules

heat lost by metal= mass of metal*specific heat of metal* change in temperature= -1585.3

=47*Cp*(32.2-100)=-1585.3, Cp =specific heat of metal

Cp=0.49 J/gm.deg.c