If 268.2 g sample of iron is placed in 1.298 L of water initially at 19.047 o C

Question

If 268.2 g sample of iron is placed in 1.298 L of water initially at 19.047oC and both substances reach a final temperature of 29.229oC, what is the initial temperature (in oC) of the metal? Assume all heat is transferred between the water and the iron If 268.2 g sample of iron is placed in 1.298 L of water initially at 19.047oC and both substances reach a final temperature of 29.229oC, what is the initial temperature (in oC) of the metal? Assume all heat is transferred between the water and the iron

Explanation / Answer

heat gained by water = heat lost by iron

=> 1.298*1000 * 1 *( 29.229 - 19.047) = 268.2 * 0.108 * ( T - 29.229)

=> T = 485.5 deg C

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