Given: latent heat of fusion of water 3.33x10^5 J/kg. A 147 g cube of ice at 0 degress C is dropped into 1.3 kg of water that was originally at 79 degrees C. What is the final temperature of the water after the ice melts.
147 g of ice, if it melts entirely, will absorb 0.147*[3.33*10^5 + 4.18 T] Joules
when heating up to temperature T in the liquid state. The water that was originaly liquid will lose heat equal to
1.3*4.18*(79 - T) J.
Set the two equal and solve for T.
Tf=65.99 degress celsius
To solve this problem, we need to find out how much heat is required to melt the ice, and then determine how this affects the temperature of the water.
Step 1: Calculate the heat required to melt the ice:
The heat required to melt the ice can be calculated using the formula:
Q = m * L_f
Where:
Q is the heat required (in Joules)
m is the mass of the ice (in kg)
L_f is the latent heat of fusion (in J/kg)
Given that the mass of the ice is 147 g (or 0.147 kg) and the latent heat of fusion of water is 3.33x10^5 J/kg, we can calculate:
Q = 0.147 kg * 3.33x10^5 J/kg
Q = 49011 J (rounded to 3 significant figures)
Step 2: Calculate the change in temperature of the water due to the heat gained from the melted ice:
The heat gained by the water when the ice melts is equal to the heat lost by the ice. This can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained or lost (in Joules)
m is the mass of the substance (in kg)
c is the specific heat capacity (in J/kg•°C)
ΔT is the change in temperature (in °C)
Given that the mass of the water is 1.3 kg and its initial temperature is 79 °C, we can calculate the change in temperature using the formula:
Q = m * c * ΔT
Rearranging the formula to solve for ΔT:
ΔT = Q / (m * c)
Where:
ΔT is the change in temperature (in °C)
Q is the heat gained or lost (in Joules)
m is the mass of the substance (in kg)
c is the specific heat capacity (in J/kg•°C)
c for water is approximately 4186 J/kg•°C
ΔT = 49011 J / (1.3 kg * 4186 J/kg•°C)
ΔT ≈ 29.3 °C (rounded to 3 significant figures)
Step 3: Calculate the final temperature of the water:
To find the final temperature of the water, add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 79 °C + 29.3 °C
Final temperature ≈ 108.3 °C (rounded to 1 decimal place)
Therefore, the final temperature of the water after the ice melts is approximately 108.3 °C.
To find the final temperature of the water after the ice melts, we can use the concept of heat transfer. The heat gained by the water must be equal to the heat lost by the ice in order for them to reach thermal equilibrium.
1. Calculate the heat lost by the ice:
The equation for heat transfer during the melting of ice is:
Q = m × Lf
where Q is the heat lost, m is the mass of the ice, and Lf is the latent heat of fusion.
Given values:
mass of the ice (m) = 147 g = 0.147 kg
latent heat of fusion (Lf) = 3.33 × 10^5 J/kg
Q = 0.147 kg × (3.33 × 10^5 J/kg) = 4.891 × 10^4 J
2. Calculate the heat gained by the water:
The equation for heat transfer is:
Q = mcΔT
where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given values:
mass of water (m) = 1.3 kg
initial temperature of water = 79°C
final temperature of water = ?
Let's assume that the final temperature of water is T.
Q = (1.3 kg) × c × (T - 79°C)
3. Equate the heat gained and lost:
Set the heat lost by the ice equal to the heat gained by the water.
4.891 × 10^4 J = (1.3 kg) × c × (T - 79°C)
4. Solve for T:
Rearrange the equation and solve for T:
(T - 79°C) = (4.891 × 10^4 J) / [(1.3 kg) × c]
T = (4.891 × 10^4 J) / [(1.3 kg) × c] + 79°C
Since the specific heat capacity of water (c) is approximately 4186 J/kg°C, you can substitute this value into the equation to find the final temperature of the water.