We should split the problem into three parts.
1) Amount of heat necessary to bring the ice from [tex]T=-18^{\circ} C[/tex] to [tex]T=0^{\circ} C[/tex]. This is given by:
[tex]Q=m C_{ice} \Delta T[/tex]
where [tex]m=76 g=0.076 kg[/tex] is the mass, [tex]\Delta T=18 ^{\circ} C=18 K[/tex] is the variation of temperature, and [tex]C_{ice} = 2.06 kJ/(Kg K)[/tex] is the specific heat of ice. Calculating, we get
[tex]Q=(0.076 kg)(2.06 kJ/(kg K))(18 K)=2.82 kJ[/tex]
2) When the ice is at [tex]T=0^{\circ} C[/tex], the heat added at this point does not change the temperature of the ice, because it is used to fuse it into water. The amount of heat needed to cause the complete fusion of ice is
[tex]Q=m L[/tex]
where [tex]L=334 kJ/kg[/tex] is the latent heat of fusion of ice. So,
[tex]Q=(0.076 kg)(334 kJ/kg)=25.38 kJ[/tex]
3) Now the ice is transformed into water. We have to bring it to [tex]T=25^{\circ} C[/tex], so the variation of temperature is [tex]\Delta T=25-0=25 ^{\circ} C=25 K[/tex]. The amount of heat needed to bring the water at this temperature is
[tex]Q=m C_{water} \Delta T[/tex]
where [tex]C_{water} = 4.186 kJ/(kg K)[/tex] is the specific heat of water. Therefore,
[tex]Q=(0.076 kg)(4.186 kJ/(kg K))(25 K)=7.95 kJ[/tex]
4) So, the total heat needed for the entire process is:
[tex]Q_{tot}=2.82 kJ + 25.38 kJ+7.95 kJ=36.15 kJ[/tex]
