joules to melt a 525g ice cube at 0∘C and to warm the liquid
and to warm the liquid HOW MUCH>
To determine the amount of energy (in joules) required to melt a 525g ice cube at 0°C and then warm the resulting liquid, we need to calculate the energy needed for each step separately and then add them together.
Step 1: Melting the ice cube
To calculate the energy required to melt the ice cube, we use the heat of fusion of water, which is the amount of energy required to convert 1 gram of ice at 0°C to 1 gram of water at 0°C. The heat of fusion for water is approximately 334 joules per gram.
Energy to melt the ice cube = mass of ice cube * heat of fusion of water
= 525g * 334 J/g
Step 2: Warming the liquid
After the ice cube has melted, we need to warm the liquid water to a desired temperature. To calculate the energy required for this step, we use the specific heat capacity of water, which is the amount of energy required to raise the temperature of 1 gram of water by 1°C. The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius.
Energy to warm the liquid = mass of liquid water * specific heat capacity of water * change in temperature
In this case, the mass of the liquid water is equal to the mass of the ice cube (since it has melted) and the change in temperature is the desired final temperature minus 0°C.
Energy to warm the liquid = 525g * 4.18 J/g°C * (final temperature - 0°C)
To determine the total energy required, you need to add the energy from step 1 (melting the ice cube) to the energy from step 2 (warming the liquid):
Total energy = energy to melt the ice cube + energy to warm the liquid
Simply substitute the values and calculations into the respective formulas to get the final answer expressed in joules.