A 314 gram sample of water at 67.5 degree Celsius loses 9.60e+03 J of heat. Calculate the final temperature of the water. ___ degrees Celcius
q = mass of water specific heat of H2O(l) T
then T = q / ( mass of water specific heat of H2O)
Substitute value in above equation
T = 9600 / (314 4.184) = 7.30
water lose heat therefore it temprature decreases by 7.30 0C
Final tempreture of water = 67.5 - 7.30 = 60.2 0C
Get Answers For Free
Most questions answered within 1 hours.