To solve this problem, we can use the principle of conservation of energy, which states that the heat lost by the hot water is equal to the heat gained by the ice.
The heat lost by the hot water can be calculated using the formula:
Q1 = mcΔT
where m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (85°C - 24°C = 61°C).
Q1 = 68g * 4.18 J/g°C * 61°C
Q1 = 17272.88 J
The heat gained by the ice can be calculated using the formula:
Q2 = mlf
where m is the mass of the ice, and lf is the heat of fusion of ice (334 J/g).
Q2 = m * 334 J/g
Since the final temperature of the system is the melting point of ice (0°C), the heat gained by the ice is also equal to the heat lost by the hot water.
Q1 = Q2
17272.88 J = m * 334 J/g
m = 17272.88 J / 334 J/g
m = 51.74 g
Therefore, the mass of the ice was 51.74 grams.
A sample of ice at -12 is placed into 68 g of water at 85C. If the final temperature of the system is 24C, what was the mass of the ice?
1 answer