An aluminum plate of mass 0.1 kg at 95oC is immersed in 0.5 litre of water at 20oC kept inside an insulating container and is then removed. If the temperature of the water is found to be 23oC, then the temperature of the aluminum plate is ___________ oC. (The specific heat of water and aluminum are 4200 J/kg-K and 900 J/kg-K respectively, the density of water is 1000 kg/m3)

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

Q = mcΔT

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

First, let's calculate the heat transferred to the water:

Qwater = mwater * cwater * ΔTwater

The mass of water is given as 0.5 litre, which is equal to 0.5 kg since the density of water is 1000 kg/m³.

The change in temperature of the water is given as (23 °C - 20 °C).

Now let's calculate the heat transferred to the aluminum plate:

Qaluminum = maluminum * caluminum * ΔTaluminum

The mass of the aluminum plate is given as 0.1 kg.

The change in temperature of the aluminum plate is what we need to find.

Given that the heat lost by the aluminum plate is equal to the heat gained by the water (Qaluminum = -Qwater), we can set up the equation:

maluminum * caluminum * ΔTaluminum = -mwater * cwater * ΔTwater

Now we can substitute the known values:

0.1 kg * 900 J/kg-K * ΔTaluminum = -0.5 kg * 4200 J/kg-K * (23 °C - 20 °C)

Simplifying the equation:

ΔTaluminum = (-0.5 kg * 4200 J/kg-K * (23 °C - 20 °C)) / (0.1 kg * 900 J/kg-K)

Calculating the value:

ΔTaluminum = (-0.5 kg * 4200 J/kg-K * 3 °C) / (0.1 kg * 900 J/kg-K)

ΔTaluminum = (-6300 J/°C) / (900 J/°C)

ΔTaluminum = -7 °C

So, the temperature of the aluminum plate would be 95 °C - 7 °C = 88 °C.

To find the temperature of the aluminum plate, we can use the principle of energy conservation. The heat lost by the aluminum plate is equal to the heat gained by the water.

Let's break down the steps to find the temperature of the aluminum plate:

Step 1: Calculate the heat lost by the aluminum plate.
The heat lost by an object can be found using the formula:
Q = m * c * ΔT
where:
Q = Heat lost or gained (in Joules)
m = mass of the object (in kg)
c = specific heat capacity of the object (in J/kg-K)
ΔT = change in temperature (in K)

Given:
Mass of the aluminum plate (m) = 0.1 kg
Specific heat capacity of aluminum (c) = 900 J/kg-K
Initial temperature of the aluminum plate (T_aluminum_initial) = 95°C
Final temperature of the water (T_water_final) = 23°C

First, we need to convert the temperatures to Kelvin:
T_aluminum_initial = 95 + 273 = 368 K
T_water_final = 23 + 273 = 296 K

Using the formula, the heat lost by the aluminum plate (Q_aluminum) is:
Q_aluminum = m * c * ΔT_aluminum

Here, ΔT_aluminum = T_aluminum_initial - T_water_final

Step 2: Calculate the heat gained by the water.
The heat gained by the water can also be found using the formula Q = m * c * ΔT.

Given:
Volume of water (V_water) = 0.5 liter = 0.5 kg (since the density of water is 1000 kg/m³)
Specific heat capacity of water (c_water) = 4200 J/kg-K
Initial temperature of the water (T_water_initial) = 20°C

First, we need to convert the temperature to Kelvin:
T_water_initial = 20 + 273 = 293 K

Using the formula, the heat gained by the water (Q_water) is:
Q_water = m * c * ΔT_water

Here, ΔT_water = T_water_final - T_water_initial

Step 3: Calculate the temperature of the aluminum plate.
Since the heat lost by the aluminum plate is equal to the heat gained by the water, we can set up the equation:
Q_aluminum = Q_water

Substituting the formulas for Q_aluminum and Q_water, we get:
m * c * ΔT_aluminum = m * c * ΔT_water

Cancelling out mass (m) and specific heat capacity (c), we get:
ΔT_aluminum = ΔT_water

Substituting the values we calculated earlier, we have:
T_aluminum_initial - T_water_final = T_water_final - T_water_initial

Now we can solve for the final temperature of the aluminum plate (T_aluminum_final):
T_aluminum_final = T_water_final - (T_water_initial - T_aluminum_initial)

Plugging in the values, we have:
T_aluminum_final = 23 - (293 - 368) = 23 - (-75) = 98°C

Therefore, the temperature of the aluminum plate is 98°C.

23. They'll come to an equilibrium temp regardless.