An aluminum spoon at 84 C is placed in a Styrofoam cup containing 0.180 kg of water at 12 C. If the final equilibrium temperature is 21 C and no heat is lost to the cup itself or the environment, what is the mass of the aluminum spoon?
Heat lost by spoon = heat gained by water
Use that heat, the aluminum temperature drop, and the specific heat of aluminum to get the mass.
Ignore the styrafoam mass. It acts as an insulator to keep the heat inside, and weighs very little.
So do I multiply all of the numbers (heat, temp. and specific heat) to get my mass? And for the temp. should I convert it to K? Or leave it to C?
0.98*0.18*21-84 = 21-12*4.186*Ms
11.1132 = 37.674Ms
Ms = 0.29 Kg
my answer wasn't right. what am i doing wrong?
To solve this problem, we need to use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, the heat energy transferred from the aluminum spoon to the water and the cup is equal to the heat energy gained by the water and the cup.
The heat energy gained or lost by an object can be calculated using the formula:
Q = mcΔT
Where:
Q is the heat energy gained or lost by the object
m is the mass of the object
c is the specific heat capacity of the object's material
ΔT is the change in temperature (final temperature - initial temperature)
We can calculate the heat energy gained by the water and the cup using the given information:
Qwater = (0.180 kg)(4186 J/kg°C)(21°C - 12°C)
Qwater = 8466 J
Next, we need to determine the heat energy lost by the aluminum spoon. Since the water and the spoon reach the same final temperature of 21°C, we can set up the following equation:
Qspoon = -(Qwater + Qcup)
Since no heat is lost to the cup itself or the environment, the heat energy lost by the spoon is equal to the sum of the heat energy gained by the water and the cup, but with the opposite sign.
Thus,
Qspoon = -8466 J
Now, we can use the equation for heat energy to calculate the mass of the aluminum spoon:
Qspoon = mcΔTspoon
-8466 J = m(897 J/kg°C)(21°C - 84°C)
Simplifying the equation:
-8466 J = m(-897 J/kg°C)(-63°C)
Dividing both sides by -897 J/kg°C:
m = -8466 J / (-897 J/kg°C)(-63°C)
m ≈ 0.135 kg
Therefore, the mass of the aluminum spoon is approximately 0.135 kg.