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 *#"^o#C) and the heat of fusion of water is 334 J/g?

1 Answer
Aug 11, 2017

Answer:

Well, you perform an energy summation.....and get #DeltaH~=9000*J.#

Explanation:

We interrogate the process.....

#H_2O(s)stackrel(Delta_1, -20^@C...0^@C)rarrH_2O(s)stackrel(Delta_2=Delta_"fusion")rarrH_2O(l)#

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*gxx334*J*g^-1=#

#20.0*gxx334*J*g^-1=6680*J#

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