i need help finding the answer
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
The specific heat of lead is 0.0305 K cal/kg deg C. The heat absorbed will be 38x(180-26)x0.0305 = 178.5 Kcal = 746.1 K joules
Melting point is the temperature at which something melts. Like ice has a melting point of 0 Celsius. The heat of fusion is the heat energy that is absorbed (or released) when a substance melts (or freezes) an is measured in energy units, either calories or joules. Ex. One gram of ice must absorb 80 calories of heat energy to change into liquid water.
The specific heat capacity of water is 4186 joules per kilogram. That is to raise 1kg or 1 litre of water by 1 degree you will need to add 4186 joules of energy. So for 15grams over 25 degrees you will need 4186/1000*15*15 joules.
3.8 x 10^5 Joules
q(joules) = mass * specific heat * change in temperature ( 8 kg = 8000 grams ) q = (8000 grams H2O)(4.180 J/gC)(70o C - 20o C) = 1.7 X 106 joules ============
15.37684 joules
What is the total number of Joules of heat absorbed by 65.00 grams of water when the temperature of th water is raised from 25.00 C to 40.00 C
Temperature is measured in celcius.Heat is measured in joules.
I assume you mean 30o Celsius. Use this formula.q(joules) = mass * specific heat * change in temperatureq = (15 grams water)(4.180 J/gC)(40o C - 30o C)= 627 joules==========( perhaps 630 joules to be in significant figures territory )
1 joule = 2.39 X 10-4 kcal 65-30 = 35 degrees 1 kcal = 1 degree kg 35 degrees X 0.5 kg / 2.39 X 10-4 kcal/joulle = 73222 joules
15480.80
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
8.200 J
Approx 4974 Joules.
10 degrees Celsius
It is 15188 Joules.