For an approximate calculation:
specific heat capacity for water = 4.18 J/(g*degC)
(how much energy is required per gram per change in degrees C)
mass = 25g
Change in temperature = 60-10 = 50 degC
energy required = mass * change in temperature * specific heat capacity
= 25g * 50 degC * 4.18 J/(g*degC)
Specific Heat capacity( S) is the amount of heat required to raise the temperature of unit mass of substance through 10C.
Let H be heat supplied, T be temperature change, m be mass of body.
Now as per our definition
Specific Heat capacity = heat supplied/mass of body*temperature change
or S = H/T*m
Putting values into the formula we have:
S = 26kJ/250g*1670C
S = 26000J/(1/4 kg * 1670C)
S = 622.75 J/kg* 0C
Hence Specific Heat capacity of substance = 622.75 J/kg* 0C.
This is done by assuming no phase change occurs during heating and substance remained in the same state as on beginning.
Using the equation Q=MC theta 600J= 0.25*C*3 600J= 0.75*C 600J/.75=C 600J*1/4=C C=15J/Kg centigrade
2.22 J/g°C
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
3.8 x 10^5 Joules
It takes 4.186 Joules to heat one gram of water by 1-degree Celsius. 4.186 * 4000 = 16,744 Joules to heat 4 kilos of water by 1-degree. 16,744 * 70 = 1,172,080 Joules. The above assumes that one litre of water weighs exactly 1 Kilogram.
There is not enough information in this problem to answer the question. You must know the mass of the iron to find the amount of joules used to heat the iron.
539 calories per gram for heat of vaporization plus 1 cal/gram/degree C 100 degrees C - 80 degrees C = 20 degrees C (539 calories + 20 calories) X 50 kg X 1000 gm/kg = 27950000 cal = 27,950 kcal
15.37684 joules
The answer is 53,683 kJ.
8.200 J
Temperature is measured in celcius.Heat is measured in joules.
Approx 4974 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.
10 degrees Celsius
It take 4.2 Joules to raise 1 gram of water by 1 degree Celsius. Each gram of water is heated by 1.26 Joules, creating an increase in temperature of .3 degrees Celsius.
3.8 x 10^5 Joules
I will use this formula. Some conversion will be required. ( I only know specific heat iron in J/gC ) q(Joules) = mass * specific heat * change in temperature Celsius 3 kilograms cast iron = 3000 grams q = (3000 g)(0.46 J/gC)(120 C - 30 C) = 124200 Joules (1 kilojoule/1000 joules) = 124.2 kilojoules of energy needed ===========================
The change in temperature is 25 degrees Celsius, meaning it takes 22.48 joules per degree of change. The specific heat of iron is 0.449 J/g degree Celsius. This means that the mass of iron must be 50.07 grams