Question

A blacksmith drops 1.5 kg piece of iron heated to 525°C into 2 L of water with temperature of 70°C. What will be the final temperature of water? Calculate amount of evaporated water, and final temperature.

Answer 1

Answer 2

SOLUTION :

Heat flows from higher temperature object to lower temperature object when they come in contact with each other. This process continues until temperature of both the objects become equal. This is the final equilibrium temperature.

Let the final temperature be tF º C when iron metal piece  at 525º C is dropped in the water which is at 70º c.

So,

Heat given up by iron metal = Heat received by water 

=> mM * sM * (525 - tF) = mW * sW * (tF - 70)

(Where, mass m is in grams, specific heats is in J / g ºC and temperature is in º ç)

=> (1.5 * 1000) * 0.450  * (525 - tF) = 2000 * 4.18 * (tF - 70)

=> 675 * (525 - tF) = 8360 * (tF - 70)

=> 8360 tF + 675 tF = 8360*70 + 675 * 525

=> 9035 tF = 939575

=> tF = 939575 / 9035

=> 103.9928 º C = 104º C =  Final temperature of water (ANSWER).

Heat available for vaporisation 

= 2000 * 4.18 * (104 - 100) =  33440 J 

Latent heat of water for vaporisation = 2230 J / g

So, amount of water vaporised

= 33440 / 2230

= 15 g (ANSWER).

And after vaporisation final temperature of water = 100º C (ANSWER).