Suppose a block of mass of lead of mass 0.4kg at temperature 95°C is dropped in 2kg of water originally at 20°C in a container.calculate the final temperature if no heat is lost to the container and environment;

heat lost by lead is gained by water

(mass Pb)(specific heat Pb)(temp change) = (mass H2O)(specific heat H2O)(temp change)

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To calculate the final temperature when a block of lead is dropped in water, we can use the principle of conservation of energy.

The equation that relates the heat gained or lost by a substance is given by:

Q = mcΔT

Where:
Q is the heat gained or lost (in Joules)
m is the mass of the substance (in kg)
c is the specific heat capacity of the substance (in J/kg°C)
ΔT is the change in temperature (in °C)

We need to calculate the initial and final temperatures of the system (water + lead).

Given information:
Mass of lead (m1) = 0.4 kg
Initial temperature of lead (T1) = 95°C
Mass of water (m2) = 2 kg
Initial temperature of water (T2) = 20°C

Since no heat is lost to the container and environment, the heat lost by the lead is gained by the water. Therefore, we can equate the two equations:

Q_lost = Q_gained

Using the equation Q = mcΔT, we can calculate the heat lost by the lead and the heat gained by the water.

Q_lost (lead) = m1 * c1 * ΔT1
Q_gained (water) = m2 * c2 * ΔT2

Substituting the given values:
Q_lost (lead) = 0.4 * c1 * (final temperature - 95)
Q_gained (water) = 2 * c2 * (final temperature - 20)

Since the two equations are equal, we can set them equal to each other and solve for the final temperature:

0.4 * c1 * (final temperature - 95) = 2 * c2 * (final temperature - 20)

Solving this equation will give us the final temperature of the system. However, we need the specific heat capacity of lead (c1) and water (c2) to proceed with the calculation. The specific heat capacity for lead is approximately 130 J/kg°C, and for water, it is about 4186 J/kg°C. Substituting these values will allow us to solve for the final temperature.