A 500 g sample of gold heated to 971.0 deg. C is placed in a calorimeter containing 220 g of water initially at 20.0 deg. C. If no heat is lost to the surrounding environment what will the final temperature of the water be? [HINT: At the end, the sample will be in thermal equilibrium with the water.]

I used the equation ((M1C1T1) + (M2C2T2))/ ((M1C1) + (M2C2)) to get 355.397 K but it was incorrect. What am I doing wrong?

Wondering how you got Kelvin for your temperature?

you stupid

To find the final temperature of the water, you can use the principle of conservation of energy. The heat gained by the water will be equal to the heat lost by the gold sample.

First, let's determine the heat gained by the water (q1):
q1 = m1 * C1 * ΔT1
where:
m1 = mass of water = 220 g
C1 = specific heat capacity of water = 4.18 J/g°C
ΔT1 = change in temperature of water = final temperature - initial temperature = T - 20°C

Next, let's determine the heat lost by the gold sample (q2):
q2 = m2 * C2 * ΔT2
where:
m2 = mass of gold sample = 500 g
C2 = specific heat capacity of gold = 0.129 J/g°C
ΔT2 = change in temperature of gold = final temperature - initial temperature = T - 971.0°C

Now, according to the principle of conservation of energy, the heat gained by the water is equal to the heat lost by the gold sample:
q1 = q2

Substituting the given values and rearranging the equation, we get:
m1 * C1 * ΔT1 = m2 * C2 * ΔT2
(220 g) * (4.18 J/g°C) * (T - 20°C) = (500 g) * (0.129 J/g°C) * (T - 971.0°C)

Now you can solve this equation for the final temperature (T). After the evaluation, you should obtain the correct answer.