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

To calculate the amount of heat that needs to be removed from the water, you can use the formula:

Q = m * c * ΔT

Where:
Q is the heat energy in Joules (J)
m is the mass of the water in kilograms (kg)
c is the specific heat capacity of water in Joules per kilogram per Kelvin (J/kg-K)
ΔT is the change in temperature in Kelvin (K)

In this case:
m = 0.15 kg
c = 4182 J/kg-K (specific heat capacity of water)
ΔT = (95 - 25) = 70 K (change in temperature)

Plugging these values into the formula:

Q = 0.15 kg * 4182 J/kg-K * 70 K
Q = 4.4 × 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 × 10^4 Joules.