Question

A piece of lead (mass = 28.0 g) at 78.0 o C is placed in a styrofoam coffee cup containing 25.0 mL of water at 21.0 oC. Assuming that no heat is lost to the cup or the surroundings, what will the final temperature of the water be? Use 1 g/mL (exact) as the density of water. (10 points)

Answer 1

The final temperature of the water will be approximately 25.53 °C.

Step-by-step explanation:

To find the final temperature of the water, we can use the principle of conservation of energy. The heat lost by the lead will be gained by the water. The formula to calculate heat transfer is:

q = m * c * ΔT

where q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, calculate the heat lost by the lead:

q_lead = m_lead * c_lead * (T_initial - T_final)

Substituting the given values:

q_lead = 28.0 g * 0.13 J/g °C * (78.0 °C - T_final)

Next, calculate the heat gained by the water:

q_water = m_water * c_water * (T_final - T_initial)

Substituting the given values:

q_water = 25.0 mL * 1 g/mL * 4.18 J/g °C * (T_final - 21.0 °C)

Since the heat lost by the lead is equal to the heat gained by the water, we can set up an equation:

q_lead = q_water

28.0 g * 0.13 J/g °C * (78.0 °C - T_final) = 25.0 mL * 1 g/mL * 4.18 J/g °C * (T_final - 21.0 °C)

Simplifying the equation:

9.24 (78.0 - T_final) = 104.5 (T_final - 21.0)

Continuing to simplify:

717.12 - 9.24T_final = 104.5T_final - 2189.5

Combining like terms:

113.74T_final = 2906.62

Finally, solving for T_final:

T_final ≈ 25.53 °C