The diagram below represents an experiment in which a student placed a 0.030 kilogram lead cube in a beaker of water at 20°C. The water in the beaker was heated to 100°C.
What is the maximum amount of heat absorbed by the lead cube during the experiment if the specific heat of lead is 0.129 J/g * oC ?
To calculate the maximum amount of heat absorbed by the lead cube, we need to use the formula:
Q = m * c * ΔT
where Q is the amount of heat absorbed, m is the mass of the lead cube, c is the specific heat of lead, and ΔT is the change in temperature.
First, we need to calculate the change in temperature of the lead cube. We know that the water in the beaker was heated from 20°C to 100°C, so the temperature difference is:
ΔT = 100°C - 20°C = 80°C
Next, we need to convert the mass of the lead cube from kilograms to grams:
m = 0.030 kg * 1000 g/kg = 30 g
Now we can plug in the values into the formula:
Q = 30 g * 0.129 J/g * oC * 80°C
Q = 309.6 J
Therefore, the maximum amount of heat absorbed by the lead cube during the experiment is 309.6 J.