The specific heat of concrete is 0.16 kcal/kgC o , and the specific heat of water is 1.00 kcal/kgC o . This means that if 1 kg of ice and 1 kg of water each receives 1 kcal of energy
To determine the change in temperature when 1 kcal of energy is added to 1 kg of ice and 1 kg of water, we need to use the specific heat values given for ice and water.
1. For the ice:
The specific heat of ice is not provided in the information given. However, we can assume that it is negligible compared to the specific heat of water. Therefore, we can consider the heat transfer for the ice as a phase change from solid to liquid, rather than a change in temperature. This phase change is known as the heat of fusion.
The heat of fusion for ice is approximately 79.7 cal/g (or 0.797 kcal/g). So, for 1 kg of ice, the heat of fusion would be 0.797 kcal/g x 1000 g = 797 kcal.
Therefore, when 1 kcal of energy is added to 1 kg of ice, it will change the phase of the ice to water, but the temperature of the ice will remain at 0 °C.
2. For the water:
The specific heat of water is given as 1.00 kcal/kg°C. This means that it takes 1.00 kcal of energy to raise the temperature of 1 kg of water by 1°C.
So, when 1 kcal of energy is added to 1 kg of water, it will increase the temperature of the water by 1°C.
In summary, when 1 kcal of energy is added to 1 kg of ice, it will change the phase of the ice to water without changing the temperature. When 1 kcal of energy is added to 1 kg of water, it will increase the temperature of the water by 1°C.
If 1 kg of ice and 1 kg of water each receives 1 kcal of energy, let's calculate the temperature change for each substance.
For the ice:
To melt the ice, we need to bring it from -20°C (assuming the initial temperature is -20°C) to 0°C, which requires:
Q = m * c * ΔT
Q = 1 kg * 0.16 kcal/kg°C * (0°C - (-20°C))
Q = 1 kg * 0.16 kcal/kg°C * 20°C
Q = 3.2 kcal
For the water:
To increase the temperature of the water from 0°C to 100°C, we use:
Q = m * c * ΔT
Q = 1 kg * 1.00 kcal/kg°C * (100°C - 0°C)
Q = 1 kg * 1.00 kcal/kg°C * 100°C
Q = 100 kcal
So, if 1 kcal of energy is provided to both the ice and water, the ice will melt completely, and the water's temperature will increase to 100°C.