how much heat has been absorbed by 2500g of water when its temperature changes from 23c to 27c PLEASE EXPLAin

jim, dd, et al. We prefer you keep the same screen name. We can help you better if you do that.

q = heat absorbed = mass H2O x specific heat H2O x (Tfinal-Tinitial)

it wud help alot if you can plug in and show how

10000

To determine the amount of heat absorbed by a substance, we need to use the formula:

Q = mcΔT

where:
Q is the amount of heat absorbed or released by the substance
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

In this case, we're looking to find the amount of heat absorbed by 2500g of water when its temperature changes from 23°C to 27°C.

1. Determine the specific heat capacity of water:
The specific heat capacity of water is approximately 4.18 J/g°C. This means that 1 gram of water requires 4.18 joules of heat energy to increase its temperature by 1°C.

2. Calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 27°C - 23°C
ΔT = 4°C

3. Plug the values into the formula:
Q = mcΔT
Q = (2500g)(4.18 J/g°C)(4°C)
Q = 41,800 J

Therefore, 2500g of water absorbed approximately 41,800 joules of heat energy when its temperature changed from 23°C to 27°C.