How much heat is required to take 4 kg of water from 20◦C to near boiling at 99◦C?
Answer in units of J.
I got 1320.88?
4.18 x 4kg x (99-20)
I put this answer into my homework and got it wrong
specific heat of water is 4.18*10^3 Joules/gram
Well, I must say, your answer is quite hot... but not in the right way! Let's calculate it properly.
The specific heat capacity of water is approximately 4186 J/kg°C. To find the heat required, we need to consider the temperature change and the mass of water.
First, let's calculate the heat required to raise the temperature of 4 kg of water from 20°C to 100°C:
Q1 = mass × specific heat capacity × temperature change
Q1 = 4 kg × 4186 J/kg°C × (100°C - 20°C)
Q1 = 4 × 4186 × 80
Q1 = 1,340,480 J
But we're not quite there yet! We also need to consider the heat required to raise the water's temperature from 100°C to 99°C:
Q2 = mass × specific heat capacity × temperature change
Q2 = 4 kg × 4186 J/kg°C × (99°C - 100°C)
Q2 = 4 × 4186 × (-1)
Q2 = -16,744 J
Now, let's put it all together:
Total heat required = Q1 + Q2
Total heat required = 1,340,480 J - 16,744 J
Total heat required ≈ 1,323,736 J
So, it seems like the actual heat required to take 4 kg of water from 20°C to near boiling at 99°C is approximately 1,323,736 J.
To calculate the amount of heat required to raise the temperature of a substance, you need to use the specific heat capacity formula. The specific heat capacity (c) is the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius.
The specific heat capacity of water is approximately 4186 J/kg°C.
First, we need to calculate the temperature change:
ΔT = final temperature - initial temperature
ΔT = 99°C - 20°C
ΔT = 79°C
Next, we can determine the amount of heat (Q) required using the formula:
Q = mass × specific heat capacity × temperature change
Q = 4 kg × 4186 J/kg°C × 79°C
Q = 1322632 J
So, the amount of heat required to take 4 kg of water from 20°C to near boiling at 99°C is 1,322,632 joules (J).