A 250 g piece of tin at 99°C is dropped in 100 g of water at 10°C. If the final temperature of the mixture is 20°C, what is the specific heat of the tin?

To find the specific heat of the tin, we can use the principle of heat transfer, which states that the heat lost by the tin is equal to the heat gained by the water.

The heat lost or gained can be calculated using the formula:

Q = m * c * ΔT

Where:
Q = heat lost or gained
m = mass
c = specific heat
ΔT = change in temperature

For the tin:
Mass of tin, m_tin = 250 g
Initial temperature of tin, T_initial_tin = 99°C
Final temperature of mixture, T_final_mixture = 20°C

For the water:
Mass of water, m_water = 100 g
Initial temperature of water, T_initial_water = 10°C
Final temperature of mixture, T_final_mixture = 20°C

First, let's calculate the heat lost by the tin. As the tin cools down, heat is lost.

Q_lost = m_tin * c_tin * (T_initial_tin - T_final_mixture)

Substituting the values, we get:

Q_lost = 250 g * c_tin * (99°C - 20°C)

Next, let's calculate the heat gained by the water. As the water warms up, heat is gained.

Q_gained = m_water * c_water * (T_final_mixture - T_initial_water)

Substituting the values, we get:

Q_gained = 100 g * c_water * (20°C - 10°C)

According to the principle of heat transfer, the heat lost by the tin is equal to the heat gained by the water. Therefore:

Q_lost = Q_gained

250 g * c_tin * (99°C - 20°C) = 100 g * c_water * (20°C - 10°C)

Now, we can solve for the specific heat of the tin, c_tin:

c_tin = (100 g * c_water * (20°C - 10°C)) / (250 g * (99°C - 20°C))

By plugging in the known values, you can calculate the specific heat of the tin.