Question

If 264.6 g sample of iron is placed in 1.005 L of water initially at 19.884^oC and both substances reach a final temperature of 28.151^oC, what is the initial temperature (in ^oC) of the metal? Assume all heat is transferred between the water and the iron.

Answer 1

specific heat data : Iron = 0.444j/g.deg.c and water =4.18 j/g.deg.c

Since water is gaining heat , iron will have to loose heat ( because no heat is transferred between water and iron)

Heat lost by iron = heat gained by water

Heat lost by iron= mass* specific heat of ion* temperature difference = 246.4*0.444*(TI-28.151)

where Ti =initial temperature of iron

Assuming the density of water to be 1 g/cc, i.e 1000 g/L

mass of water = 1.005*1000 =1005 gm

Heat gained by water = 10005*4.18*(28.151-19.884)=34729 joules

34729= 246.4*0.444*(Ti-28.151)

Ti-28.151= 296

T= 296+28.151= 324.151 deg.c

Heat gained by water=