# How much heat is required to convert 20.0 g of ice at -50.0°C to liquid water at 0.0°C? The specific heat of ice is 2.06 J/(g *"^oC) and the heat of fusion of water is 334 J/g?

Aug 11, 2017

Well, you perform an energy summation.....and get $\Delta H \cong 9000 \cdot J .$

#### Explanation:

We interrogate the process.....

${H}_{2} O \left(s\right) \stackrel{{\Delta}_{1} , - {20}^{\circ} C \ldots {0}^{\circ} C}{\rightarrow} {H}_{2} O \left(s\right) \stackrel{{\Delta}_{2} = {\Delta}_{\text{fusion}}}{\rightarrow} {H}_{2} O \left(l\right)$

Now Delta_1=20.0*gxx2.06*J*g^-1*""^@C^-1xxDeltaT=

20.0*gxx2.06*J*g^-1*""^@C^-1xx50 ""^@C=2060*J

And ${\Delta}_{2} = 20.0 \cdot g \times 334 \cdot J \cdot {g}^{-} 1 =$

$20.0 \cdot g \times 334 \cdot J \cdot {g}^{-} 1 = 6680 \cdot J$

"Total heat"=Delta_1+Delta_2=2060*J+6680*J=??*J