How do you calculate this

a 25g ice cube at its melting point is added to calorimeter water at 25.4 C. When thermal equilibrium is reached, the final temperature of the mixture is 15C. The original volume in the calorimeter was

a) 93.4mL
b) 136mL
c) 209mL
d) 228mL

We will assume no heat is lost to the surroundings. The warm water in the calorimeter will lose heat and go down from 25.4 to 15 (Δt = 10.4 C).

The heat will do two things:
1) melt the 25 g of ice
2) raise the 25 g of liquid from 0 to 15 C (Δt = 15 C)

(x)(10.4 C)(4.184 J/g C) = [(25 g) (334 J/g)] + [(25g)(15 C)(4.184 J/g C)]

(x)(10.4 C)(4.184 J/g C) = 9919 J

x = 227.95 g of water

Since 1 g of water occupies 1 mL of volume, we pick answer choice d.

Update: The 334 J/g is the heat of fusion for ice and the 4.184 value is the specific heat for liquid water.

Ice heat of fusion (heat to melt) - 334 J/g
Specific heat of water - 4.18 J/gC
Formula - heat=(specific heat)(mass)(temperature change)

Calculate the heat required to melt the 25g ice:
Q=334*25=8350 J

Calculate the temperature of the water after the melting (we don't know the mass yet so we'll call it m), and I made the Q negative here because heat is being removed:
-8350=4.18*m*(T-25.4)
T=-2000/m + 25.4

Set up an equation and plug in the temperature found above to find the mass of the original water.
Q(added water)=-Q(original water)
4.18*25*(15-0)=-(4.18*m*(15-(-2000/m + 25.4)))
m=228 gram
Density of water is 1g/1mL, so the volume is also 228 mL, so the answer is D.