How much heat in kcal must be added to 0.32 kg of water at room temperature (20°C) to raise its temperature to 36°C?
.32 times (36-20)
q = mass x specific heat water x (Tfinal-Tinitial)
Well, let me put on my thinking hat (and red nose) for a moment. To calculate that, we need to use the formula for heat, Q = mcΔT. Where Q is the energy required in calories, m is the mass of the water in grams, c is the specific heat capacity of water (1 calorie per gram per degree Celsius), and ΔT is the change in temperature.
So, let's calculate this:
Q = (0.32 kg) * (1000 g/kg) * (1 cal/g°C) * (36°C - 20°C)
Q = 320 cal/g°C * 16°C
Q = 5120 calories!
But since you asked for the answer in kcal, we'll divide that by 1000:
Q = 5.12 kcal!
So, my friend, you need to add 5.12 kcal of heat to that water to raise its temperature to 36°C. That's hot stuff!
To calculate the amount of heat required to raise the temperature of a substance, you can use the specific heat formula:
Q = m * c * ΔT
Where:
Q = heat energy (in calories or kilocalories)
m = mass of the substance (in grams or kilograms)
c = specific heat capacity of the substance (in calories/gram °C or kilocalories/kilogram °C)
ΔT = change in temperature (in °C)
First, let's convert the mass of water from kg to grams:
0.32 kg = 320 grams
Next, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 1 calorie/gram °C or 1 kilocalorie/kilogram °C.
Now, we can plug the values into the formula:
Q = 320 g * 1 kcal/kg°C * (36°C - 20°C)
Q = 320 g * 1 kcal/kg°C * 16°C
Q = 5120 kcal
Therefore, you would need to add 5120 kilocalories of heat to raise the temperature of 0.32 kg of water from 20°C to 36°C.