if 9560 J of energy were absorbed by 25.0g of ice at 0 degrees Celsius what would be the final temperature? First you must calculate the energy required to melt ice completely. With the remaining energy (Q).

This is the equation i came up with but i'm not even sure if its the right equation.
Q= (s)(m)(T).

No.

9650=.025*Hf + .025cwater(Tf-0)
heat=heatmeltingice+heatwater.

Look up the heat of fusion (J/kg) for water, and then the specific heat of water (J/kgK)

To find the final temperature, you need to consider two processes: (1) melting the ice completely and (2) heating the resulting water to the final temperature.

First, let's calculate the energy required to melt the ice completely. The equation you mentioned, Q = (s)(m)(ΔT), is a good starting point. However, since we are dealing with the phase change from ice to water (melting), the equation should be adjusted. The specific heat capacity (s) is replaced with the heat of fusion (ΔH_f), which represents the energy required to change the phase of the substance.

The equation becomes:
Q1 = (ΔH_f)(m)

Now, let's calculate Q1:
Given:
Energy absorbed, Q absorbed = 9560 J
Mass of ice, m = 25.0 g

We also need the heat of fusion for ice (ΔH_f). The heat of fusion for ice is 334 J/g.

Calculating Q1:
Q1 = (ΔH_f)(m)
Q1 = (334 J/g)(25.0 g)
Q1 = 8,350 J

So, the energy required to melt the ice completely (Q1) is 8,350 J.

Next, since the remaining energy (Q) needs to heat the water to the final temperature, we calculate Q as follows:

Q2 = Q absorbed - Q1
Q2 = 9560 J - 8350 J
Q2 = 1,210 J

Now, we can use the equation Q = (s)(m)(ΔT) to find the change in temperature (ΔT) of the water.

Given:
Q2 = 1,210 J
Mass of water, m = 25.0 g
The specific heat capacity of water (s) is 4.18 J/g°C.

Calculating ΔT:
Q2 = (s)(m)(ΔT)
1,210 J = (4.18 J/g°C)(25.0 g)(ΔT)
ΔT = 1,210 J / (4.18 J/g°C)(25.0 g)
ΔT ≈ 11.6 °C

Therefore, the final temperature of the water would be approximately 11.6°C.