A 0.500kg block of metal with an initial temperature of 30 degrees C is dropped into a container holding 1.12kg of water at 20.0 degrees C. If the final temperature of the block-water system is 20.4 degrees C, what is the specific heat of the metal? Assume the container can be ignored, and that no heat is exchanged with the surroundings. (specific heat of water, cw= 4186 J/ kg K

To find the specific heat of the metal, we can use the principle of energy conservation. The heat gained by the metal equals the heat lost by the water.

Let's start by calculating the heat gained by the metal:

Q gained by metal = m * c * ΔT

Where:
m = mass of the metal (0.500 kg)
c = specific heat of the metal (unknown)
ΔT = change in temperature of the metal (final temperature - initial temperature)

Q gained by metal = 0.500 kg * c * (20.4 °C - 30 °C)

Next, let's calculate the heat lost by the water:

Q lost by water = m * cw * ΔT

Where:
m = mass of the water (1.12 kg)
cw = specific heat of water (4186 J/kg K)
ΔT = change in temperature of the water (final temperature - initial temperature)

Q lost by water = 1.12 kg * 4186 J/kg K * (20.4 °C - 20.0 °C)

According to the principle of energy conservation, Q gained by metal = Q lost by water. Therefore:

0.500 kg * c * (20.4 °C - 30 °C) = 1.12 kg * 4186 J/kg K * (20.4 °C - 20.0 °C)

Simplifying the equation:

0.500 kg * c * (-9.6 °C) = 1.12 kg * 4186 J/kg K * 0.4 °C

Solving for c:

c = (1.12 kg * 4186 J/kg K * 0.4 °C) / (0.500 kg * -9.6 °C)

c ≈ 872 J/kg K

Therefore, the specific heat of the metal is approximately 872 J/kg K.

To solve this problem, we can use the equation for heat transfer:

Q = mcΔT

Where Q is the heat transfer, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, let's calculate the heat transfer for the water:

Qwater = mwater * cwater * ΔTwater

We are given that the mass of the water, mwater, is 1.12 kg, the specific heat of water, cwater, is 4186 J/kg K, and the change in temperature, ΔTwater, is the final temperature (20.4 degrees C) minus the initial temperature (20.0 degrees C).

ΔTwater = 20.4 degrees C - 20.0 degrees C = 0.4 degrees C

Now we can calculate Qwater:

Qwater = 1.12 kg * 4186 J/kg K * 0.4 degrees C

Next, let's calculate the heat transfer for the metal:

Qmetal = mmetal * cmetal * ΔTmetal

We are given that the mass of the metal, mmetal, is 0.500 kg, the change in temperature, ΔTmetal, is the final temperature (20.4 degrees C) minus the initial temperature (30.0 degrees C), and cmetal is the specific heat we want to find.

ΔTmetal = 20.4 degrees C - 30.0 degrees C = -9.6 degrees C

Now we can rewrite the equation for Qmetal:

Qmetal = 0.500 kg * cmetal * -9.6 degrees C

Since no heat is exchanged with the surroundings, the total heat transfer is equal to zero:

Qtotal = Qwater + Qmetal = 0

Plugging in the values we have calculated, we can solve for cmetal:

1.12 kg * 4186 J/kg K * 0.4 degrees C + 0.500 kg * cmetal * -9.6 degrees C = 0

Simplifying the equation, we get:

4799.44 J - 4.8 cmetal = 0

Rearranging the equation to solve for cmetal, we get:

4.8 cmetal = 4799.44 J

cmetal = 4799.44 J / 4.8

cmetal ≈ 999.88 J/kg K

Therefore, the specific heat of the metal is approximately 999.88 J/kg K.