thermochemistry

Question

The temperature of 293 g of water is changed from 10.6 to 38.9 °C. How much energy was transferred into the water? Calculate your answer in joules, kilojoules, calories, and kilocalories.

Answer 1

The equation to use for heat transfer is q = mC∆T

q = heat

m = mass = 293 g

C = specific heat of water = 4.184 J/g/deg

∆T = change in temperature = 38.9 - 10.6 = 28.3 degrees

q = (293 g)(4.184 J/g/deg)(28.3 deg) = 34,693 Joules

For kilojoules: 34,693 J x 1 kJ/1000 J = 34.693 kJ

For calories (1 cal = 4.184 J): 34,693 J x 1 cal/4.184 J = 8292 cal

For kcal: 8292 cal x 1 kcal/1000 cal = 8.292 kcal