We are told that 100.0"g of H"_2"O"100.0g of H2O went from 20^@C20∘C to 25^@C25∘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.
