The enthalpy change for converting 1.00 mol of ice at -25.0 â°c to water at 70.0 â°c is kj. the specific heats of ice, water, and steam are 2

Answer is: 6,16 kJ.
1) changing  temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 2 J/g·°C · 25°C 
Q₁ = 900 J.
m(H₂O) = 1mol · 18 g/mol = 18 g.
C - specific heat of ice.
2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C 
Q₁ = 5266,8 J.
C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.