A piece of aluminum (specific heat 910 J/kg C) of mass 200 g at 80C is dropped into a styrofoam cup filled with 100 mL of water at 20C. What the final temperatures of the water and the aluminum? Show calculations.

I have Q=m c T
but I got 80C which I know is wrong because I set up Q1=Q2 which gave me that, could you please help me set up this problem.

Call final temp = T

Heat lost by Al = specific heat AL * mass of Al * (80-T)

Heat gained by water - specific heat of H2O * mass of H2O *(T-20)

Then
Heat lost by Al = Heat gained by H2O

right but I got 80 C both times, that cannot possibly be correct.

To solve this problem, we need to consider the heat transfer between the aluminum and the water.

Step 1: Calculate the heat gained or lost by the aluminum:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum

Given:
m_aluminum = 200 g
c_aluminum = 910 J/kg°C
ΔT_aluminum = final temperature - initial temperature

Since the final temperature is unknown, we can leave it as T_aluminum.

Step 2: Calculate the heat gained or lost by the water:
Q_water = m_water * c_water * ΔT_water

Given:
m_water = 100 mL (which is equivalent to 100 g, since 1 mL of water has a mass of 1 g)
c_water = 4186 J/kg°C (specific heat of water)
ΔT_water = final temperature - initial temperature

Since the final temperature is unknown, we can leave it as T_water.

Step 3: Equate the heat gained by the aluminum to the heat gained by the water:
Q_aluminum = Q_water

m_aluminum * c_aluminum * ΔT_aluminum = m_water * c_water * ΔT_water

Step 4: Plug in the given values and solve for the final temperatures:
200 g * 910 J/kg°C * (T_aluminum - 80°C) = 100 g * 4186 J/kg°C * (T_water - 20°C)

Now, let's solve for T_aluminum and T_water.

200 * 910 * T_aluminum - 200 * 910 * 80 = 100 * 4186 * T_water - 100 * 4186 * 20

182,000 * T_aluminum - 145,600 * 200 = 418,600 * T_water - 83,720 * 100

182,000 * T_aluminum - 29,120,000 = 418,600 * T_water - 8,372,000

182,000 * T_aluminum - 418,600 * T_water = -8,372,000 + 29,120,000

-236,600 * T_water + 236,600 * T_aluminum = 20,748,000

Divide through by 236,600:

T_water - T_aluminum = 87.667

Since we don't know the final temperatures, we'll leave them as T_water and T_aluminum.

So, the final equations are:
T_water - T_aluminum = 87.667
T_water + T_aluminum = 100

Solve this system of equations to find the final temperatures.

To solve this problem, you need to apply the principle of conservation of energy, which states that the energy lost by the aluminum will be gained by the water in the cup. Let's break down the steps to solve this:

Step 1: Calculate the energy gained or lost by the aluminum.
The equation Q = mcΔT represents the energy gained or lost by an object, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Given:
Mass of aluminum (m1) = 200 g = 0.2 kg
Specific heat capacity of aluminum (c1) = 910 J/kg°C
Initial temperature of aluminum (T1) = 80°C
Final temperature of aluminum (Tf)

Using the formula:
Q1 = m1 * c1 * ΔT1

For the aluminum, the change in temperature (ΔT1) is calculated as:
ΔT1 = Tf - T1

Step 2: Calculate the energy gained or lost by the water.
The equation Q = mcΔT represents the energy gained or lost by an object, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Given:
Mass of water (m2) = 100 mL = 0.1 kg (since 1 mL of water has a mass of 1 g)
Specific heat capacity of water (c2) = 4186 J/kg°C (specific heat of water)
Initial temperature of water (T2) = 20°C
Final temperature of water (Tf)

Using the formula:
Q2 = m2 * c2 * ΔT2

For the water, the change in temperature (ΔT2) is calculated as:
ΔT2 = Tf - T2

Step 3: Equate the energy gained by the water and the energy lost by the aluminum.
Q1 = Q2

m1 * c1 * ΔT1 = m2 * c2 * ΔT2

Substitute the values we have:
0.2 kg * 910 J/kg°C * (Tf - 80°C) = 0.1 kg * 4186 J/kg°C * (Tf - 20°C)

Step 4: Solve for Tf (the final temperature).
Let's solve this equation for Tf:

0.2 kg * 910 J/kg°C * Tf - 0.2 kg * 910 J/kg°C * 80°C = 0.1 kg * 4186 J/kg°C * Tf - 0.1 kg * 4186 J/kg°C * 20°C

Simplifying further:
182 J/°C * Tf - 14560 J/°C = 418.6 J/°C * Tf - 8372 J/°C

Combine like terms:
182 J/°C * Tf - 418.6 J/°C * Tf = 14560 J/°C - 8372 J/°C

Multiply by -1 to simplify:
-236.6 J/°C * Tf = -6192 J/°C

Divide by -236.6 J/°C to isolate Tf:
Tf = (-6192 J/°C) / (-236.6 J/°C)

Tf ≈ 26.16°C (rounded to two decimal places)

The final temperature of the water and the aluminum is approximately 26.16°C.