How many kilojoules of heat are absorbed when 1.00L of water is heated from 18 degrees C to 85 degrees C?
Did I do this right?
1000g H2O x 4.18 x 67 = 280060 kilojoules
(my book says the specific heat for water in J/g x degrees C = 4.18, and for cal/g x degrees C = 1.00) Would i multiply 1000g x 67 x by 4.18 or 1?
Whether you use 1 cal/g*C or 4.18 J/g*C depends upon what unit you want the answer in. The first one gives the answer in calories, the second in joules. Your math is ok but the answer is not kJ, but J. You need to divide you answer by 1000 to convert to kJ.
To calculate the amount of heat absorbed when heating water, you can use the equation:
q = m * c * ΔT
where:
- q is the amount of heat absorbed or released (in joules)
- m is the mass of the water (in grams)
- c is the specific heat capacity of water (in J/g°C)
- ΔT is the change in temperature (in °C)
In your case, you have 1.00 L of water, but you need to convert this to grams since the c value is given in J/g°C. The density of water is approximately 1 g/mL, so 1.00 L of water is equal to 1000 grams.
Given that the specific heat for water is 4.18 J/g°C, you would multiply the mass (1000 g) by the change in temperature (67 °C) and by the specific heat (4.18 J/g°C).
Calculating the value:
q = 1000 g * 4.18 J/g°C * 67°C
q ≈ 279,860 J
To convert the result to kilojoules (kJ), you need to divide by 1000:
q ≈ 279.86 kJ
So, it seems that your initial calculation of 280,060 kJ is almost correct. There might be a very small arithmetic error in your calculation. The correct amount of heat absorbed when 1.00 L of water is heated from 18°C to 85°C is approximately 279.86 kJ.