We are told that #100.0"g of H"_2"O"# went from #20^@C# to #25^@C#. The reference Specific Heat gives us the equation for how much heat energy was absorbed by the water:

#Q= cmDeltaT" [1]"#

where #c= "the specific heat of water" = (4.139"J")/("g H"_2"O"*"C"^@)#, #m = "the mass" = 100.0"g of H"_2"O"#, and #DeltaT = "the change in temperature" = 25^@C-20^@C#

Substitute these values into equation [1]:

#Q= (4.139"J")/("g H"_2"O"*"C"^@)(100.0"g H"_2"O")(25^@C-20^@C)#

The water absorbed:

#Q = 2069.5"J"#

of heat energy.

The energy came from the metal and it obeys the same equation, equation [1]:

#Q= cmDeltaT" [1]"#

We know everything in this equation, except the value of, c -- the specific heat for the metal, therefore, we can substitute in the known values and then solve for c:

We know that #Q = -2069.5"J"#, because the energy that went into the water came out from the hot metal:

#-2069.5"J"= cmDeltaT" [1.1]"#

We know that #m = 30"g of Metal"#

#-2069.5"J"= c(30"g of Metal")DeltaT" [1.2]"#

We know that the temperature of the metal went from #110^@"C"# to the same temperature as the water, #25^@"C"#, therefore, #DeltaT = 25^@"C"-110^@"C"#:

#-2069.5"J"= c(30"g of Metal")(25^@"C"-110^@"C")" [1.3]"#

To solve for c, we divide by the mass and the change in temperature:

#c =(-2069.5"J")/((30"g of Metal")(25^@"C"-110^@"C")")#

#c = (0.812"J")/("g of Metal"*C^@)#

This is the specific heat of the metal.