Assume all temperatures to be exact.
An aluminum spoon at 84 degree C is placed in a Styrofoam cup containing 0.180 kg of water at 12 degree C. If the final equilibrium temperature is 21 degree C and no heat is lost to the cup itself or the environment, what is the mass of the aluminum spoon?
Please help!
To solve this problem, we need to apply the principle of conservation of energy, specifically the law of heat exchange.
The heat gained by the water will be equal to the heat lost by the aluminum spoon. The equation we will use is:
Qwater = -Qspoon
where Qwater is the heat gained by the water and Qspoon is the heat lost by the aluminum spoon.
The equation for heat gained or lost (Q) is given by:
Q = mcΔT
where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the water, we can calculate the heat gained as follows:
Qwater = mwater * cwater * ΔTwater
For the aluminum spoon, we can calculate the heat lost as follows:
Qspoon = mspoon * cspoon * ΔTspoon
Since no heat is lost to the cup or the environment, the heat gained by the water (Qwater) will be equal to the heat lost by the spoon (Qspoon):
mwater * cwater * ΔTwater = - mspoon * cspoon * ΔTspoon
Now, let's plug in the values given in the problem:
mwater = 0.180 kg (mass of water)
cwater = 4186 J/kg°C (specific heat capacity of water)
ΔTwater = (21 - 12) °C (change in temperature of water)
cspoon = 900 J/kg°C (specific heat capacity of aluminum)
ΔTspoon = (21 - 84) °C (change in temperature of spoon)
Simplifying the equation, we get:
0.180 * 4186 * (21 - 12) = - mspoon * 900 * (21 - 84)
Now, we can solve for the mass of the spoon (mspoon):
mspoon = (0.180 * 4186 * (21 - 12)) / (900 * (21 - 84))
mspoon ≈ 0.046 kg (rounded to three decimal places)
Therefore, the mass of the aluminum spoon is approximately 0.046 kg.