A 1000 g sample of metal with a specific heat of 0.50 J/goC is heated to 100.0oC and then placed in a 50.0 g sample of water at 20.0oC. (Hint: Specific heat of water at 20.0oC is 4.182 J/goC). What is the final temperature of the metal and the water? Write your answer to the correct number of significant figures in scientific notation.

heat gained by water equals heat lost by metal

50.0 * (t - 20.0) * 4.182 =
... 1000 * (100.0 - t) * 0.50

To solve this problem, we can use the principle of conservation of energy. The energy gained by the metal will be equal to the energy lost by the water.

First, let's calculate the energy gained by the metal. We can use the equation:

Qmetal = mcΔT

Where:
Qmetal is the heat gained by the metal,
m is the mass of the metal (1000 g),
c is the specific heat of the metal (0.50 J/goC),
and ΔT is the change in temperature of the metal (final temperature - initial temperature).

We know that the initial temperature of the metal is 100.0oC, and we need to find the final temperature.

Now, let's calculate the energy lost by the water. We can use the equation:

Qwater = mcΔT

Where:
Qwater is the heat lost by the water,
m is the mass of the water (50.0 g),
c is the specific heat of water (4.182 J/goC),
and ΔT is the change in temperature of the water (final temperature - initial temperature).

The initial temperature of the water is 20.0oC, and we need to find the final temperature.

Since the total energy gained by the metal is equal to the total energy lost by the water, we can set up the equation:

Qmetal = Qwater

mcΔT (metal) = mcΔT (water)

Substituting the given values:

1000 g * 0.50 J/goC * (Tfinal - 100.0oC) = 50.0 g * 4.182 J/goC * (Tfinal - 20.0oC)

Simplifying the equation:

500 * (Tfinal - 100.0) = 209.1 * (Tfinal - 20.0)

500Tfinal - 50000 = 209.1Tfinal - 4182

290.9Tfinal = 45818

Tfinal ≈ 157.6oC

Therefore, the final temperature of both the metal and the water is approximately 157.6oC.

To find the final temperature of the metal and water, we can use the principle of conservation of energy. The heat transferred from the metal to the water is equal to the heat gained by the water.

We can calculate the heat transferred using the equation:

Q = mcΔT

Where:
Q is the heat transferred
m is the mass
c is the specific heat
ΔT is the change in temperature

For the metal:
m = 1000 g
c = 0.50 J/goC
ΔT = final temperature - initial temperature = final temperature - 100.0oC

For the water:
m = 50.0 g
c = 4.182 J/goC
ΔT = final temperature - initial temperature = final temperature - 20.0oC

Since the heat transferred from the metal to the water is equal, we can set up an equation:

Qmetal = Qwater

Using the equation Q = mcΔT, we can write:

mmetal * cmetal * ΔTmetal = mwater * cwater * ΔTwater

Substituting the given values:

1000g * 0.50J/goC * (final temperature - 100.0oC) = 50.0g * 4.182J/goC * (final temperature - 20.0oC)

Simplifying the equation:

500J * (final temperature - 100.0oC) = 209.1J * (final temperature - 20.0oC)

Expanding and rearranging the equation:

500J * final temperature - 50000J = 209.1J * final temperature - 4182J

Combining like terms:

290.9J * final temperature = 45818J

Dividing both sides of the equation by 290.9J:

final temperature = 45818J / 290.9J

final temperature = 157.6oC

Therefore, the final temperature of both the metal and water is approximately 157.6oC, written in scientific notation as 1.58 x 10^2oC.