how much heat enrgy must be taken in 24 gram of water to reduce the temperature of 7 degrres celcius
To reduce temperature, heat energy must be removed, not "taken in".
Do you mean that you want to reduce the temperature BY 7 degrees? To reduce a temperature "OF 7 degrees" doesn't make sernse. Reduce it to what?
In this case, the amount that must be removed is
M*C*(deltaT) = 24*7 = 168 calories
since the specific heat C of water is 1.00
To determine how much heat energy is required to reduce the temperature of water, you can use the equation:
Q = m * c * ΔT
Where:
Q is the heat energy (in joules),
m is the mass of water (in grams),
c is the specific heat capacity of water (approximately 4.18 J/g°C),
ΔT is the change in temperature (in degrees Celsius).
In this case, you have:
m = 24 grams (mass of water)
ΔT = -7°C (change in temperature, negative because the temperature is reducing)
Now, you can simply substitute the values into the formula:
Q = 24 g * 4.18 J/g°C * (-7°C)
Calculating the right side of the equation:
Q = -706.08 J
Therefore, you would need to take in approximately -706.08 joules (remember the negative sign indicates that heat is being removed) of energy to reduce the temperature of 24 grams of water by 7 degrees Celsius.