Question

A piece of iron (400 g) is heated in a flame to 330°C and then plunged into a beaker containing 1000g of water at 20°C. What is the final equilibrium temperature of the water? The specific heat of iron is .451J/g•C°

specific heat of water = 4.184 J/g•C°

Answer 1

This problem can be summarized thusly:

q_{lost by iron} = q_{gained by water} + q_{gained by calorimeter}

therefore :

(400 g) (330^oC - x) (0.451 J/g.C^o) = (1000 g) (x - 20^oC) (4.184 j/g.C^o)

(330^oC - x) / (x - 20^oC) = (1000 * 4.184 ) / (400 * 0.451)

(330^oC - x) / (x - 20^oC) = 23.193

= (330^oC - x) = 23.193 (x - 20^oC)

   (330^oC - x) = 23.193 x - 463.86^oC

= 793.86^oC = 24.193 x

=> x = 793.86^oC / 24.193

=> x = 32.81 ^oC approximately