Eighty calories must be added to one gram of ice at 0ºC to melt it. What is the approximate mechanical equivalent of this amount of energy?
The answer is 80 calories.
Want it in Joules? 80cal*1j/.239cal
To determine the approximate mechanical equivalent of the energy required to melt one gram of ice, we need to understand the concept of specific heat.
Specific heat is the amount of heat energy required to raise the temperature of a substance by a certain amount. The specific heat of ice is approximately 0.5 calories/gram/°C. This means that it takes 0.5 calories to raise the temperature of one gram of ice by one degree Celsius.
In this case, we need to add 80 calories to melt one gram of ice at 0ºC. However, this energy is not used to raise the temperature, but rather to change the state of the substance from solid to liquid. This is known as the latent heat of fusion.
The latent heat of fusion of ice is approximately 80 calories/gram. This means that 80 calories are required to melt one gram of ice at 0ºC without raising its temperature.
Now, to find the mechanical equivalent of this energy, we can use the concept of mechanical equivalent of heat, which relates heat energy to mechanical work.
The mechanical equivalent of heat is the amount of mechanical work needed to produce one calorie of heat energy. It is denoted by the symbol J and is approximately equal to 4.186 Joules/calorie.
So, to find the mechanical equivalent of 80 calories, we multiply it by the mechanical equivalent of heat:
80 calories * 4.186 J/calorie ≈ 334.88 Joules
Therefore, the approximate mechanical equivalent of the energy required to melt one gram of ice at 0ºC is approximately 334.88 Joules.