How much heat must be removed from 0.15 kg of water to cool it from 95 d/C to 25 d/C? The specific heat capasity of water is 4182 J/kg-k.
mC(T1-T2)
0.15(4182)(70)
= 4.4 x 10^4 J
Thinking is correct. I didn't check calc work.
Well, you've got the right formula and calculations there! I guess you could say that removing that much heat from water is a pretty chilling job. But hey, it's nothing water couldn't handle! Keep up the cool calculations!
To calculate the amount of heat that must be removed from the water, you can use the formula:
Q = mC(T1 - T2)
where:
Q is the amount of heat (in joules)
m is the mass of water (in kilograms)
C is the specific heat capacity of water (in J/kg-K)
T1 is the initial temperature (in degrees Celsius)
T2 is the final temperature (in degrees Celsius)
Given:
m = 0.15 kg
C = 4182 J/kg-K
T1 = 95 °C (convert to Kelvin: 95 + 273 = 368 K)
T2 = 25 °C (convert to Kelvin: 25 + 273 = 298 K)
Plugging the given values into the formula, we get:
Q = (0.15 kg)(4182 J/kg-K)(368 K - 298 K)
Q = (0.15)(4182)(70)
Q = 4.4 × 10^4 J
Therefore, the amount of heat that must be removed from the water is 4.4 × 10^4 joules.
To calculate the amount of heat that needs to be removed from water to cool it down, we can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat transferred (in joules).
m is the mass of the substance (in kilograms).
c is the specific heat capacity of the substance (in joules per kilogram-kelvin).
ΔT is the change in temperature (in kelvin).
In this case, we have:
m = 0.15 kg (mass of water)
c = 4182 J/kg-K (specific heat capacity of water)
ΔT = (T1 - T2) = (95°C - 25°C) = 70°C
Now, let's substitute the values into the formula:
Q = 0.15 kg * 4182 J/kg-K * 70 K
Q = 4.4 x 10^4 J
Therefore, the amount of heat that must be removed from 0.15 kg of water to cool it from 95°C to 25°C is 4.4 x 10^4 J.