Question

a) how many joules are needed to heat up 325 grams of water from 25.1 degrees c to 43.8 degrees c? how many calories? b) how many joules are needed to heat up 325 grams of aluminum from 25.1 degrees c to 89.5 degrees c? how many calories? the c for aluminum is = 0.897 j/g degrees c

Answer 1

a) Mass of water, m= 325g Specificheat of liquid water, s = 4.184 J/g.°C Change in temperature AT = 7; - = 43.8°C-25.1°C = 18.7°C Amount of heat absorbed by water, grater = mSAT = 325 g * 4.184 * 18.7°C = 25428J = 2.54x10*1 404 ... 1 cal com Amount of heat absorbedin calories, enter = 2.54x10* JX-404 = 6071 cal b) Mass of aluminum, m= 325g Specificheat of aluminum , s = 0.897 J/g °C Change in temperature AT = 7; -7 = 89.5°C-25.1°C = 64.4°C Amount of heat absorbed by water, grafer = mSAT = 325 g x 4.184 * 64.4°C = 87571J = 8.76x10* I 10+ 1 cal Amount of heat absorbedin calories, qanter = 8.76x10* Jx0 20020 = 20930 cal