A 24-g ice cube floats in 180 g of water in a 100-g copper cup; all are at a temperature of 0°C. A piece of lead at 92°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33 105 J/kg and 4,186 J/kg · °C, respectively. The specific heat of lead and copper are 128 and 387 J/kg · °C, respectively.)

http://answers.yahoo.com/question/index?qid=20100629105308AAL8TBp

How did they get 9.804?

Thank you

To solve this problem, we can use the principle of conservation of energy. The heat lost by the lead when it cools down is equal to the heat gained by the ice, water, and cup to bring them to an equilibrium temperature.

The heat lost by the lead can be calculated using the specific heat formula:

Q_lead = m_lead * c_lead * (T_final - T_initial)

where Q_lead is the heat lost by the lead, m_lead is the mass of the lead, c_lead is the specific heat of lead, T_final is the final temperature, and T_initial is the initial temperature.

The heat gained by the ice can be calculated using the heat of fusion formula:

Q_ice = m_ice * L_fusion

where Q_ice is the heat gained by the ice, m_ice is the mass of the ice, and L_fusion is the heat of fusion.

The heat gained by the water can be calculated using the specific heat formula:

Q_water = m_water * c_water * (T_final - T_initial)

where Q_water is the heat gained by the water, m_water is the mass of the water, c_water is the specific heat of water, T_final is the final temperature, and T_initial is the initial temperature.

The heat gained by the cup can be calculated using the specific heat formula:

Q_cup = m_cup * c_cup * (T_final - T_initial)

where Q_cup is the heat gained by the cup, m_cup is the mass of the cup, c_cup is the specific heat of the cup, T_final is the final temperature, and T_initial is the initial temperature.

Since the cup does not change its temperature significantly, we can ignore its heat gain:

Q_cup = 0

Since the system is in equilibrium, the heat lost by the lead is equal to the heat gained by the ice and water:

Q_lead = Q_ice + Q_water

Now, let's plug in the given values and solve for the mass of the lead, m_lead:

m_lead * c_lead * (T_final - T_initial) = m_ice * L_fusion + m_water * c_water * (T_final - T_initial)

Substituting the given values:

m_lead * 128 * (12 - 92) = 24 * (3.33 * 10^5) + 180 * 4186 * (12 - 0)

Simplifying the equation:

-1152m_lead = 7.992 * 10^6 + 9.00488 * 10^6

Combining the terms:

-1152m_lead = 16.99788 * 10^6

Dividing by -1152:

m_lead = (16.99788 * 10^6) / -1152

Calculating the answer:

m_lead = -14743.88 g

Since mass cannot be negative, we take the absolute value:

m_lead = 14743.88 g

Thus, the mass of the lead is approximately 14743.88 g.