How much heat is contained in 100 kg of water at 60.0 °C?
Be mindful of significant figure rules for both multiplication and addition.
30 kcal
2.10 x 104 kcal
2.40 x 104 kcal
2.77 x 104 kcal
You have to understand three different components of this questions. H20 is ice from -273 to 0C and water from 0 - 60C. You also have to figure in the latency heat of ice turning to water. (1ookg)(.5)(273) + (100kg)(1)(60) + (100)(80) = 2.77x10^4
I had this question too, you helped me a lot
To determine the amount of heat contained in 100 kg of water at 60.0 °C, we need to use the formula:
Q = mcΔT
Where:
Q is the amount of heat (in calories or joules)
m is the mass of the substance (in kilograms)
c is the specific heat capacity of the substance (in calories per gram per degree Celsius or in joules per gram per Kelvin)
ΔT is the change in temperature (in degrees Celsius or Kelvin)
In this case, the mass of water is 100 kg, the specific heat capacity of water is approximately 1 calorie per gram per degree Celsius (or 4184 joules per gram per Kelvin), and the change in temperature is 60.0 °C.
First, convert the mass from kilograms to grams:
100 kg = 100,000 grams
Next, substitute the values into the formula:
Q = (100,000 grams) x (1 calorie/gram/°C) x (60.0 °C)
Simplifying:
Q = 6,000,000 calories
To convert calories to kilocalories, divide by 1000:
Q = 6,000 kilocalories
Rounding to the appropriate number of significant figures (based on the given options), we obtain:
Q = 6.00 x 10^3 kcal
Therefore, the correct answer is 2.40 x 10^4 kcal.