Final answer:
To determine the minimum power rating of the heater, we calculate the total energy required to heat both the coffee mug and water from 25° C to 100° C using their masses and specific heat capacities. After calculating the total energy, we divide it by the time (120 seconds) to find the power. This results in the minimum power rating needed for the heater to bring the coffee mug and water to boiling in two minutes.
Step-by-step explanation:
To find the minimum power rating of the heater, we first need to determine the total energy required to bring the water to boiling from the initial temperature. This can be calculated using the equation q = mcΔT, where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We have a coffee mug and water with a combined mass of 0.22 kg + 0.34 kg and a specific heat capacity of 960 J/(kg · C°) for the mug and 4200 J/(kg · C°) for water respectively. The initial temperature is 25° C and we want to bring it to the boiling point of water which is 100° C.
The total heat energy required is the sum of energies needed to heat up both the water and the mug:
- Energy for water: q_water = (0.34 kg)(4200 J/(kg · C°))(100° C - 25° C)
- Energy for mug: q_mug = (0.22 kg)(960 J/(kg · C°))(100° C - 25° C)
Once we calculate q_water and q_mug, we add both to get the total energy required. The power is then found by dividing the total energy by the time (in seconds) it takes to boil the water, which gives:
Power = Total energy / time
Since power is energy per unit time, and we know the time taken is two minutes, or 120 seconds, we just plug in the values to find the minimum power rating of the heater.