a 5.0 x 10^1g piece of a metal in thermal equilibrium with boiling water is removed and placed in 1.0 x 10^2g of a cooling water bath (specific heat= 1.00cal/g degrees celsius at 20.0 degrees Celsius . The final temperature of metal and the water bath is 28.0 degrees celsius What is the specific heat of the metal?
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for specific heat
metal.
To find the specific heat of the metal, we can use the principle of energy conservation. The heat lost by the metal when it cools down is equal to the heat gained by the cooling water bath. The formula to calculate the heat is given by:
Q = m * c * ΔT
Where:
Q = Heat (calories)
m = Mass of the substance (grams)
c = Specific heat capacity (cal/g °C)
ΔT = Change in temperature (°C)
In this case, we know the mass and specific heat capacity of the cooling water bath, the initial and final temperatures of both the metal and the cooling water bath. We'll use this information to calculate the specific heat of the metal.
First, we calculate the heat lost by the metal. Considering that the metal is initially in thermal equilibrium with boiling water, its initial temperature is 100°C. The final temperature is given as 28°C. Therefore, the change in temperature (ΔT) for the metal is:
ΔT = 28°C - 100°C = -72°C
Since the metal is cooling down, the negative sign indicates that the temperature is decreasing. Now, let's calculate the heat lost by the metal:
Q (metal) = m (metal) * c (metal) * ΔT (metal)
Next, we calculate the heat gained by the cooling water bath. We know the specific heat capacity of the water bath is 1.00 cal/g°C. The mass of the water bath is 100g. The initial temperature is 20°C, and the final temperature is 28°C. The change in temperature (ΔT) for the water bath is:
ΔT = 28°C - 20°C = 8°C
Using this information, we can calculate the heat gained by the water bath:
Q (water) = m (water) * c (water) * ΔT (water)
Since energy is conserved, the heat lost by the metal is equal to the heat gained by the water bath:
Q (metal) = Q (water)
m (metal) * c (metal) * ΔT (metal) = m (water) * c (water) * ΔT (water)
Now, we can rearrange this equation to solve for the specific heat of the metal, c (metal):
c (metal) = [m (water) * c (water) * ΔT (water)] / [m (metal) * ΔT (metal)]
Substituting the known values:
c (metal) = [100g * 1.00 cal/g°C * 8°C] / [5.0 x 10^1g * -72°C]
Simplifying the equation:
c (metal) = 800 cal / (-3600 g°C)
Finally, we can find the specific heat of the metal:
c (metal) = -0.222 cal/g°C
Therefore, the specific heat of the metal is approximately -0.222 cal/g°C.