How many kg of ice needs to be added to 1.31 kg of water at 64.9°C to cool the water to 14.9°C when the Latent heat of ice = 80 kcal/kg?
m₁cΔt₁=λm₂+m₂cΔt₂
m₂=m₁cΔt₁/(λ+cΔt₂) =
=1.31•4218•(64.9-14.9)/(336000+4218•14.9)=
=0.69 kg
(Latent heat of ice = 80 kcal/kg=336000 J/kg)
To solve this problem, we can use the principle of energy conservation, which states that the heat lost by the water will be gained by the ice. The equation is:
Heat gained by ice = Heat lost by water
First, we need to calculate the heat lost by water. We can use the specific heat capacity formula:
Q = m * c * ΔT
where:
Q = heat lost by water (in calories)
m = mass of water (in grams)
c = specific heat capacity of water (1 calorie/gram°C)
ΔT = change in temperature (in °C)
Given:
Mass of water (m) = 1.31 kg = 1310 grams
Initial temperature (T1) = 64.9°C
Final temperature (T2) = 14.9°C
Calculating the heat lost by water:
Q = m * c * ΔT
Q = 1310g * 1 cal/g°C * (14.9°C - 64.9°C)
Q = 1310g * 1 cal/g°C * (-50°C)
Q = -65500 calories
Since the water is losing heat, the value is negative.
Next, we will calculate the heat gained by the ice. We can use the formula:
Q = m * Latent heat of ice
Given:
Latent heat of ice = 80 kcal/kg
Calculating the heat gained by the ice:
Q = m * Latent heat of ice
-65500 cal = m * 80 kcal/kg
To convert calories to kilocalories:
1 kcal = 1000 cal
-65500 cal = m * (80 kcal/kg * 1000 cal/kcal)
-65500 cal = 80000 m cal
Solving for m (mass of ice):
m = (-65500 cal) / 80000 kcal/kg
m = -0.81875 kg
Since mass cannot be negative, we take the absolute value:
m = 0.81875 kg
Therefore, approximately 0.819 kg (or 819 grams) of ice needs to be added to cool the water from 64.9°C to 14.9°C.
To determine how many kilograms of ice need to be added to cool the water, we need to calculate the heat transfer required to lower the temperature.
Step 1: Calculate the heat lost by the water.
The formula to calculate heat transfer is: Q = m × C × ΔT
Where:
Q = heat transfer (in kcal)
m = mass (in kg)
C = specific heat capacity (in kcal/kg°C)
ΔT = change in temperature (in °C)
Given:
m_water = 1.31 kg (mass of water)
C_water = 1 kcal/kg°C (specific heat capacity of water)
ΔT_water = 64.9°C - 14.9°C = 50°C (change in temperature of water)
Using the formula, we can calculate the heat lost by the water:
Q_water = m_water × C_water × ΔT_water
Q_water = 1.31 kg × 1 kcal/kg°C × 50°C
Q_water = 65.5 kcal
Step 2: Calculate the heat gained by the ice.
The heat gained by the ice will be equal to the heat lost by the water since they are at the same temperature.
Q_ice = Q_water
Step 3: Calculate the mass of ice required.
The formula to calculate heat transfer using latent heat is: Q = m × L
Where:
Q = heat transfer (in kcal)
m = mass (in kg)
L = latent heat (in kcal/kg)
Given:
L_ice = 80 kcal/kg (latent heat of ice)
Q_ice = 65.5 kcal (heat gained by the ice)
Using the formula, we can calculate the mass of ice required:
Q_ice = m_ice × L_ice
m_ice = Q_ice / L_ice
m_ice = 65.5 kcal / 80 kcal/kg
m_ice ≈ 0.819 kg
Therefore, approximately 0.819 kg of ice needs to be added to 1.31 kg of water at 64.9°C to cool the water to 14.9°C when the latent heat of ice is 80 kcal/kg.