what is the heat in joules required to melt 25 grams of ice?
heat fusion is 6.01 kJ/mol. mols = 25/18 = about 1.38
Then 1.38 x 6.01 = ? kJ. That answer above is VERY close.
To calculate the heat required to melt 25 grams of ice, we need to use the specific heat capacity and heat of fusion of ice.
1. First, let's find the heat required to raise the temperature of the ice from its initial temperature to its melting point. The specific heat capacity of ice is 2.09 J/g°C.
Heat = mass * specific heat capacity * change in temperature
The initial temperature of ice is 0°C, and the melting point of ice is 0°C. So the change in temperature is 0°C - 0°C = 0°C.
Heat = 25 g * 2.09 J/g°C * 0°C
= 0 J
Therefore, no heat is required to raise the temperature of the ice to its melting point since its initial temperature is already at the melting point.
2. Next, we need to find the heat required to melt the ice. The heat of fusion of ice is 334 J/g.
Heat = mass * heat of fusion
Heat = 25 g * 334 J/g
= 8350 J
Therefore, the heat required to melt 25 grams of ice is 8350 Joules (J).
To determine the heat required to melt 25 grams of ice, we need to know the specific heat capacity and the heat of fusion of ice.
1. Specific Heat Capacity: The specific heat capacity (c) of a substance is the heat energy required to raise the temperature of 1 gram of that substance by 1 degree Celsius. For water, the specific heat capacity is approximately 4.184 J/g°C.
2. Heat of Fusion: The heat of fusion (ΔHfus) is the amount of heat energy required to convert 1 gram of a substance from its solid state to its liquid state at its melting point. For water, the heat of fusion is approximately 334 J/g.
Now that we have the necessary information, let's calculate the heat required to melt 25 grams of ice.
Heat = Mass × Heat of Fusion
Heat = 25 g × 334 J/g
Heat = 8350 J
Therefore, it would require approximately 8350 joules of heat energy to melt 25 grams of ice.