Answer to Question #296230 in General Chemistry for golden

Question

How much heat is released when 358 g of water cools from 68.3°C to 18.9°C? The specific heat of water is 4.184 J/g°C.

Answer 1

2 phase changes will occur in this process : from steam to water and then water to ice.

The heat released (q) = m . c. delta T

where m is mass in grams

c is the specific heat

delta T is change in tempertaure

So here :

q1 = 24.0 x 2.16 x (104 - 100)

= 207.36 J

q2 = mass x heat of vaporization

= 24.0 x 2260

= 54240 J

q3 = 24.0 x 4.184 x (100-0)

= 10041.6 J

q4 = mass x heat of fusion

= 24.0 x 334

= 8016 J

q5 = 24.0 x 2.06 x [0-(-3.03)]

= 149.8 J

So the total heat released = q1 + q2 + q3 + q4 + q5

= 72654.76 J

So heat released in calories = 72654.76 / 4.184

= 17364.9 cal