How much ice (in grams) with the temperature of 0∘C would have to melt to lower the temperature of 355mL of water from 26∘C to 4∘C? (Assume the density of water is 1.0 g/mL.)
To solve this problem, we need to determine the amount of heat that needs to be transferred from the water to the ice in order to lower the temperature.
First, we can calculate the heat absorbed by the water using the formula:
Q = m * C * ΔT
where Q is the heat absorbed by the water, m is the mass of water, C is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
- Density of water = 1.0 g/mL
- Volume of water = 355 mL
- Initial temperature of water = 26°C
- Final temperature of water = 4°C
To find the mass of water, we can use the formula:
mass = volume * density
mass = 355 mL * 1.0 g/mL
mass = 355 g
Next, we can substitute the values into the formula to calculate the heat absorbed by the water:
Q = 355 g * 4.18 J/g°C * (4°C - 26°C)
Q = 355 g * 4.18 J/g°C * (-22°C)
Q = -32569.6 J (Note: The negative sign indicates that heat is being lost by the water)
To find the amount of ice that needs to melt to absorb this amount of heat, we can use the heat of fusion of ice. The heat of fusion of ice is the amount of heat required to melt ice at 0°C.
The heat of fusion of ice is 334 J/g. Therefore, to find the amount of ice that needs to melt, we can rearrange the formula:
Q = m * heat of fusion
m = Q / heat of fusion
m = -32569.6 J / 334 J/g
m ≈ -97.42 g (Note: The negative sign indicates a decrease in mass due to melting of the ice)
Since mass cannot be negative, we can conclude that approximately 97.42 grams of ice would need to melt to lower the temperature of the water from 26°C to 4°C.
To solve this problem, we need to determine the amount of heat transferred from the water to the ice as the ice melts.
Given:
Volume of water (Vw) = 355 mL
Initial temperature of water (Tw) = 26°C
Final temperature of water (Tf) = 4°C
Density of water (ρw) = 1.0 g/mL
Step 1: Calculate the mass of water (mw)
Density of water = mass / volume
mw = ρw * Vw
= 1.0 g/mL * 355 mL
= 355 g
Step 2: Calculate the heat transferred from the water.
Heat transferred (Q) = mass of water * specific heat capacity * change in temperature
Specific heat capacity of water (Cw) = 4.18 J/g°C (approx)
Q = mw * Cw * (Tw - Tf)
= 355 g * 4.18 J/g°C * (26°C - 4°C)
= 355 g * 4.18 J/g°C * 22°C
= 32529.4 J
Step 3: Calculate the heat required to melt the ice.
Heat required to melt the ice (Qm) = mass of ice * heat of fusion
Heat of fusion of ice (Hf) = 334 J/g
Qm = Q = mw * Hf
Mass of ice (mi) = Qm / Hf
= 32529.4 J / 334 J/g
≈ 97.39 g
Therefore, approximately 97.39 grams of ice with a temperature of 0°C would have to melt to lower the temperature of 355 mL of water from 26°C to 4°C.
(mass ice x heat fusion) + [mass ice x specific heat water x (Tfinal-Tinitial)] [mass H2O x specific heat H2O x (Tfinal-Tionitial)] = 0
Substitute and solve for mass ice. Tfinal is 4, Tinitial is 26, mass H2O (that's the 26 C H2O) is 355. Massice is the only unknown.