Answer
We can start by using the conservation of energy principle, where the energy lost by the iceberg is equal to the energy gained by the water in the lake:
$Q_{lost} = -Q_{gained}$
where $Q$ is the amount of energy transferred, which can be calculated using the specific heat capacity of water and the mass and temperature change of the water:
$Q_{gained} = m_w c_w \Delta T$
where $m_w$ is the mass of water, $c_w$ is the specific heat capacity of water, and $\Delta T$ is the change in temperature of the water. Since the iceberg is initially colder than the water, we can assume that all of the energy lost by the iceberg is used to heat up the water.
The energy lost by the iceberg can be calculated as:
$Q_{lost} = m_i c_i \Delta T$
where $m_i$ is the mass of the iceberg, $c_i$ is the specific heat capacity of ice, and $\Delta T$ is the change in temperature of the iceberg. Since the iceberg is assumed to be at a constant temperature of -10.0°C, $\Delta T$ is equal to the final temperature of the water minus -10.0°C.
Equating these two equations and solving for the final temperature of the water, $T_f$, we get:
$m_w c_w \Delta T = m_i c_i (T_f - (-10.0°C))$
$T_f = \frac{m_w c_w \Delta T}{m_i c_i} - 10.0°C$
Substituting the given values, we get:
$m_w$ = 185,000,000 g
$c_w$ = 4.18 J/(g*K)
$c_i$ = 2.01 J/(g*K)
$m_i$ = 17.3 * 10^6 g
$\Delta T$ = $T_f - 6.40°C$
Solving for $T_f$, we get:
$T_f$ = -1.03°C
The final temperature of the water is -1.03°C.
To determine if any ice remains, we need to check if the final temperature of the water is less than or equal to the freezing point of water, which is 0°C. Since -1.03°C is less than 0°C, some of the water will freeze. The mass of ice that forms can be calculated using the heat of fusion of water:
$Q_{gained} = m_{ice} L_f$
where $m_{ice}$ is the mass of ice that forms and $L_f$ is the heat of fusion of water, which is 334 J/g.
Substituting the given values, we get:
$m_{ice} = \frac{m_w c_w \Delta T}{L_f}$
$m_{ice} = \frac{(185,000,000 g) (4.18 J/(g*K)) (-1.03°C - 6.40°C)}{334 J/g}$
$m_{ice} = 15,783,832 g$
Therefore, 15,783,832 g or approximately 15.8 metric tons of ice will form.
Work Step by Step
We can start by using the conservation of energy principle, where the energy lost by the iceberg is equal to the energy gained by the water in the lake:
$Q_{lost} = -Q_{gained}$
where $Q$ is the amount of energy transferred, which can be calculated using the specific heat capacity of water and the mass and temperature change of the water:
$Q_{gained} = m_w c_w \Delta T$
where $m_w$ is the mass of water, $c_w$ is the specific heat capacity of water, and $\Delta T$ is the change in temperature of the water. Since the iceberg is initially colder than the water, we can assume that all of the energy lost by the iceberg is used to heat up the water.
The energy lost by the iceberg can be calculated as:
$Q_{lost} = m_i c_i \Delta T$
where $m_i$ is the mass of the iceberg, $c_i$ is the specific heat capacity of ice, and $\Delta T$ is the change in temperature of the iceberg. Since the iceberg is assumed to be at a constant temperature of -10.0°C, $\Delta T$ is equal to the final temperature of the water minus -10.0°C.
Equating these two equations and solving for the final temperature of the water, $T_f$, we get:
$m_w c_w \Delta T = m_i c_i (T_f - (-10.0°C))$
$T_f = \frac{m_w c_w \Delta T}{m_i c_i} - 10.0°C$
Substituting the given values, we get:
$m_w$ = 185,000,000 g
$c_w$ = 4.18 J/(g*K)
$c_i$ = 2.01 J/(g*K)
$m_i$ = 17.3 * 10^6 g
$\Delta T$ = $T_f - 6.40°C$
Solving for $T_f$, we get:
$T_f$ = -1.03°C
The final temperature of the water is -1.03°C.
To determine if any ice remains, we need to check if the final temperature of the water is less than or equal to the freezing point of water, which is 0°C. Since -1.03°C is less than 0°C, some of the water will freeze. The mass of ice that forms can be calculated using the heat of fusion of water:
$Q_{gained} = m_{ice} L_f$
where $m_{ice}$ is the mass of ice that forms and $L_f$ is the heat of fusion of water, which is 334 J/g.
Substituting the given values, we get:
$m_{ice} = \frac{m_w c_w \Delta T}{L_f}$
$m_{ice} = \frac{(185,000,000 g) (4.18 J/(g*K)) (-1.03°C - 6.40°C)}{334 J/g}$
$m_{ice} = 15,783,832 g$
Therefore, 15,783,832 g or approximately 15.8 metric tons of ice will form.
