How many calories of heat are required to change 1 gram of ice at 0°C to liquid water at 0°C?
calories=mass*specificheat=1g*80cal/g=80cal
To determine the amount of heat energy required to change 1 gram of ice at 0°C to liquid water at 0°C, we need to consider the specific heat capacity and latent heat of fusion.
Step 1: Calculate the heat required to raise the temperature of ice from 0°C to its melting point at 0°C.
The specific heat capacity of ice is approximately 2.09 J/g°C. We can use the formula: Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q = 1g * 2.09 J/g°C * (0°C - 0°C) = 0 J
Therefore, no heat is required to raise the temperature of ice from 0°C to its melting point at 0°C.
Step 2: Calculate the heat required for the phase change from ice to water at 0°C.
The latent heat of fusion for ice is 334 J/g. We can use the formula: Q = m * L, where Q is the heat energy, m is the mass, and L is the latent heat of fusion.
Q = 1g * 334 J/g = 334 J
Therefore, 334 calories of heat are required to change 1 gram of ice at 0°C to liquid water at 0°C.
To determine the amount of calories of heat required to change ice at 0°C to liquid water at 0°C, we need to consider the specific heat capacity and the heat of fusion of water.
First, we need to calculate the heat required to change the temperature of the ice from 0°C to 0°C. The specific heat capacity of ice is 0.5 calories/gram °C.
The formula to calculate the heat required to change the temperature is:
Q = m * c * ΔT
where:
Q is the heat energy
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
In this case, the mass of the ice is 1 gram, the specific heat capacity is 0.5 calories/gram °C, and the change in temperature is 0°C. Plugging these values into the formula, we get:
Q = 1g * 0.5 cal/g°C * 0°C
Q = 0 calories
Since there is no change in temperature, no calorie of heat is required.
However, to fully convert ice at 0°C to liquid water at 0°C, we need to consider the heat of fusion of water. The heat of fusion for water is 79.7 calories/gram. This value represents the amount of heat energy required to change 1 gram of solid ice at 0°C to liquid water at 0°C without raising the temperature.
So, to find the total calories of heat required, we add the heat of fusion to the previously calculated heat energy:
Total Q = Q (change in temperature) + Heat of fusion
Total Q = 0 calories + 79.7 calories
Total Q = 79.7 calories
Therefore, it requires 79.7 calories of heat to change 1 gram of ice at 0°C to liquid water at 0°C.