.You have an ice cube that is 150 g and is at a temperature of 0°C. You drop the ice into
water that is initially at 15ºC and the ice melts. As a result, the water cools of by exactly 1ºC to
become 14ºC. How much water was there? In other words, what mass of water at 15ºC can be
cooled by 1ºC by removal of the heat needed to melt 150 g of ice at 0ºC?
I think you have two problems here. The first one ends with "was there?" Then the second one begins. The first one must take into account the fact that the 150 g ice melts at zero C and that water must be raised to 14 C while the other water is being reduced by 1 C (from 15 C to 14 C). The second problem ignores the 150 g H2O produced by the melted ice.
Are you working in J or calories? Calories?
150 g x heat fusion (80 cal/g) = 12,000 calories.
If you want the water to go from 15 C to 14 C, then you can remove 12,000 calories. Heat capacity of H2O is 1 cal/g*C, that means 12,000 g H2O. That ignores the 150 g H2O at zero that comes from the melted ice and that must be raised to 14 C. That could remove another 150 x 14 = 2100.
The two answers are 12,000 g H2O ignoring the melted ice or 14,100 g H2O including the ice.
To determine the mass of water needed to cool down by 1ºC by removing the heat needed to melt 150 g of ice at 0ºC, we can use the principle of heat transfer.
The heat lost by the water is equal to the heat gained by the ice during the melting process. The formula for 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, let's find out the heat gained by the ice during melting. The heat needed to melt the ice can be calculated using the formula: q_ice = m_ice * ΔH_fusion, where q_ice is the heat gained by the ice, m_ice is the mass of the ice, and ΔH_fusion is the enthalpy of fusion.
The enthalpy of fusion (ΔH_fusion) for water is 334 J/g.
q_ice = m_ice * ΔH_fusion
q_ice = 150 g * 334 J/g
q_ice = 50100 J
Now, let's calculate the heat lost by the water. We know that the change in temperature (ΔT) is 1ºC.
q_water = m_water * c_water * ΔT
q_water = m_water * 4.186 J/g°C * 1ºC
q_water = 4.186 m_water J
Since the heat gained by the ice is equal to the heat lost by the water, we can set up the equation:
q_ice = q_water
50100 J = 4.186 m_water
Simplifying the equation, we can solve for m_water:
m_water = 50100 J / 4.186 J/g
m_water ≈ 11985 g
Therefore, approximately 11985 grams (or 11.985 kg) of water at 15ºC can be cooled by 1ºC by removing the heat needed to melt 150 g of ice at 0ºC.