A piece of metal of mass 20.0g @ 100°C is placed in a calorimeter containing 50.7g of water at 22.0°C. The final temperature of the mixture is 25.7°C. What is the specific heat capacity of the metal? Assume that no energy leaves to the surroundings.

Not sure how to work this problem out. Help please? Thank u!

[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0

You have all except specific heat metal. Solve for that.

If 46.1 g Cu at 11.6 C is placed in 85.0 g H2O at 72.4 C, what is the final temperature of the mixture? (specific heat capacities: Cu: 0.385 J/gK Water: 4.184 J/gK)?

To solve this problem, we can use the principle of energy conservation. The energy lost by the metal is gained by the water in the calorimeter.

Step 1: Calculate the energy lost by the metal.
The energy lost by the metal can be calculated using the formula:

Q = m * c * ΔT,

where Q is the energy lost, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature of the metal.

Given:
m (mass of the metal) = 20.0g,
ΔT (change in temperature of the metal) = (25.7°C - 100°C) = -74.3°C.

Step 2: Calculate the energy gained by the water.
The energy gained by the water can be calculated using the formula:

Q = m * c * ΔT,

where Q is the energy gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.

Given:
m (mass of the water) = 50.7g,
ΔT (change in temperature of the water) = (25.7°C - 22.0°C) = 3.7°C.

Step 3: Equate the energy lost by the metal to the energy gained by the water.
Since no energy is lost to the surroundings, the energy lost by the metal is equal to the energy gained by the water.
So, we can set up the equation:

m * c (metal) * ΔT (metal) = m * c (water) * ΔT (water).

Substituting the given values, we have:

(20.0g) * c (metal) * (-74.3°C) = (50.7g) * c (water) * (3.7°C).

Step 4: Solve for the specific heat capacity of the metal.
To find the specific heat capacity of the metal (c metal), we rearrange the equation from Step 3 and solve for c metal:

c (metal) = [(50.7g) * c (water) * (3.7°C)] / [(20.0g) * (-74.3°C)].

Now, calculate the specific heat capacity of the metal using the given values and the equation above.

To find the specific heat capacity of the metal, you can use the principle of energy conservation. The formula used for this is:

Q = mcΔT

Where:
Q is the amount of heat energy transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.

In this problem, the heat gained by the metal will be equal to the heat lost by the water. Since there is no energy lost to the surroundings, the amount of heat gained by the metal can be calculated as:

Q_gain = mcΔT

Similarly, the amount of heat lost by the water can be calculated as:

Q_loss = mcΔT

Since Q_gain = Q_loss, we can substitute in the values provided in the problem:

m_metal * c_metal * ΔT_metal = m_water * c_water * ΔT_water

To find the specific heat capacity of the metal, we rearrange the equation:

c_metal = (m_water * c_water * ΔT_water) / (m_metal * ΔT_metal)

Substituting the values from the problem:

m_metal = 20.0 g
m_water = 50.7 g
c_water = 4.184 J/g·°C (specific heat capacity of water)
ΔT_water = 25.7°C - 22.0°C = 3.7°C
ΔT_metal = 25.7°C - 100°C = -74.3°C

Now, plug in these values to calculate the specific heat capacity of the metal:

c_metal = (50.7 g * 4.184 J/g·°C * 3.7°C) / (20.0 g * -74.3°C)

Simplifying the equation, we get:

c_metal ≈ -9.77 J/g·°C

Note: The negative sign indicates that the metal has a lower specific heat capacity compared to water, which means it heats up and cools down faster than water.