How much ice (in grams) would have to melt to lower the temperature of 350 of water from 24 to 4? (Assume the density of water is 1.0 .)

Assuming density of water = 1,

specific heat = 1 cal/g and
initial temperature of ice = 0°C
latent heat of fusion of ice
mass of ice = x
mass of water = 350

calories above 0° before mixing:
E0 = 350*24+x*0+x*(-80)
calories after mixing:
E1 = (350+x)*4
Equating calories before & after mixing:
350*24-80x=(350+x)*4
Solve for x to get x=250/3 g

To calculate the amount of ice that would have to melt to lower the temperature of water from 24 to 4 degrees Celsius, we need to use the concept of specific heat and the latent heat of fusion.

1. Calculate the heat lost by the water:
Heat lost = Mass of water * Specific heat of water * Change in temperature
Mass of water = Volume of water * Density of water
Density of water = 1.0 g/mL

Given:
Volume of water = 350 mL
Specific heat of water = 4.18 J/g°C (at constant pressure)
Change in temperature = 24°C - 4°C = 20°C

Mass of water = 350 mL * 1.0 g/mL = 350 g
Heat lost = 350 g * 4.18 J/g°C * 20°C (Equation 1)

2. Calculate the heat gained by the ice:
Heat gained = Mass of ice * Latent heat of fusion
Latent heat of fusion of ice = 334 J/g

Let's assume that all the ice melts to water at 0°C and then it warms up to 4°C. So, the change in temperature for ice = 4°C.

Heat gained = Mass of ice * Latent heat of fusion + Mass of ice * Specific heat of water * Change in temperature (Equation 2)

3. At the phase transition (melting/freezing), the heat lost by the water is equal to the heat gained by the ice:
Heat lost = Heat gained
350 g * 4.18 J/g°C * 20°C = Mass of ice * 334 J/g + Mass of ice * 4.18 J/g°C * 4°C (Equation 3)

4. Solve for the mass of ice:
Rearrange Equation 3:
350 g * 4.18 J/g°C * 20°C = Mass of ice * (334 J/g + 4.18 J/g°C * 4°C)
350 g * 4.18 J/g°C * 20°C = Mass of ice * (334 J/g + 16.72 J/g)

Simplify:
350 g * 4.18 J/g°C * 20°C = Mass of ice * 350.72 J/g
Mass of ice = (350 g * 4.18 J/g°C * 20°C) / 350.72 J/g

Mass of ice ≈ 39.29 g

Therefore, approximately 39.29 grams of ice would have to melt to lower the temperature of 350 grams of water from 24 to 4 degrees Celsius.

To solve this problem, we need to use the equation:

\(q = mcΔT\)

where:
- \(q\) is the heat transferred (in Joules)
- \(m\) is the mass of the substance (in grams)
- \(c\) is the specific heat capacity of the substance (in J/g°C)
- \(ΔT\) is the change in temperature (in °C)

In this case, we want to find the mass of ice that would have to melt to lower the temperature of water from 24°C to 4°C.

1. Calculate the heat transferred (\(q\)) using the equation above:
In this case, the water is losing heat, so \(q\) will be negative.
Since the specific heat capacity of water is 4.18 J/g°C:
\(q = (-350 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (-20°C)\)

2. Calculate the heat required to melt the ice (\(q_melt\)):
\(q_melt = m_{ice} \times H_f\)
\(H_f\) is the heat of fusion (amount of heat required to melt 1 gram of a substance)
For water, \(H_f\) is 334 J/g.

Since the ice is gaining heat, \(q_melt\) will be positive.
Rearranging the equation, we find:
\(m_{ice} = \frac{q_melt}{H_f}\)

3. Substitute the \(q_melt\) value into the equation:
\(m_{ice} = \frac{q}{H_f}\)

4. Calculate the \(q_melt\) value using the \(q\) value calculated in step 1:
\(-\frac{q}{H_f} = \frac{(-350 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (-20°C)}{334 \, \text{J/g}}\)

Simplifying the equation will yield the mass of ice required to lower the temperature of the water.