How much heat energy is needed to convert 125.0g of ice from -27.0ºC to a liquid at 50.0ºC?
To calculate the amount of heat energy needed to convert ice to a liquid at a different temperature, you need to consider two steps: first, raising the temperature of the ice from its initial temperature (-27.0ºC) to its melting point (0ºC), and second, heating the water from 0ºC to the desired final temperature (50.0ºC).
Let's break down the calculation for each step:
Step 1: Raising the ice from -27.0ºC to 0ºC
To calculate the heat energy needed in this step, you can use the formula:
q = m * C * ΔT
where:
q is the heat energy,
m is the mass of the substance (ice in this case),
C is the specific heat capacity of the substance,
ΔT is the change in temperature.
For ice, the specific heat capacity is approximately 2.09 J/gºC.
So, the heat energy needed in this step can be calculated as follows:
q = (mass of ice) * (specific heat capacity of ice) * (change in temperature)
q = 125.0 g * 2.09 J/gºC * (0ºC - (-27.0ºC))
Step 2: Heating the water from 0ºC to 50.0ºC
To calculate the heat energy needed in this step, you can use the same formula as before, but now considering the specific heat capacity for water, which is approximately 4.18 J/gºC.
Again, the heat energy needed can be calculated as follows:
q = (mass of water) * (specific heat capacity of water) * (change in temperature)
q = (mass of water) * 4.18 J/gºC * (50.0ºC - 0ºC)
Finally, you add up the heat energy required for both steps to get the total heat energy needed.