a beaker contains 500g of water at a temperature of 80 C. How much ice at a temperature -20 C must be dropped in the water so that the final temperature of the system will be 50 C
the sume of heats gained is zero.
Heat to heat ice from -20 to 0
heat to melt ice at zero
heat to heat melted to temp from 0 to 50
heat to heat original water from 80 to 50
add these up, and find mass ice. Notice one of the heats above is negative, so it makes the heats add to zero...
To find the amount of ice needed to reach the final temperature, we need to consider the heat gained by the ice and the heat lost by the water.
First, let's calculate the heat lost by the water.
The specific heat capacity of water is 4.18 J/g·°C.
The initial temperature of the water is 80°C, and the final temperature is 50°C.
The formula to calculate the heat lost is:
Heat lost = mass × specific heat capacity × change in temperature
Given:
Mass of water = 500 g
Specific heat capacity of water = 4.18 J/g·°C
Change in temperature = 80°C - 50°C = 30°C
Heat lost = 500 g × 4.18 J/g·°C × 30°C = 62700 J
Now, let's calculate the heat gained by the ice.
The heat required to raise the temperature of ice from -20°C to its melting point (0°C) is calculated using the formula:
Heat gained = mass × specific heat capacity × change in temperature
Given:
Initial temperature of ice = -20°C
Final temperature of ice = 0°C
Specific heat capacity of ice = 2.09 J/g·°C
Change in temperature = 0°C - (-20°C) = 20°C
Heat gained = mass × 2.09 J/g·°C × 20°C = 41.8 J × mass
To achieve thermal equilibrium, the heat gained by the ice must be equal to the heat lost by the water. Therefore, we can set up an equation:
62700 J = 41.8 J × mass
Solving for mass:
mass = 62700 J / 41.8 J ≈ 1499 g
So, approximately 1499 grams of ice at a temperature of -20°C must be dropped into the water to reach a final temperature of 50°C.