Final answer:
The final temperature of the water after absorbing 300 calories of heat is 45°C. This is calculated using the formula Q = mcΔT and rearranging it to find the change in temperature ΔT, which when added to the initial temperature gives us the final temperature.
Step-by-step explanation:
To find the final temperature of the water after absorbing 300 calories of heat, we use the formula for heat absorption, Q = mcΔT, where Q is the heat absorbed by the substance in calories, m is the mass of the water in grams, c is the specific heat capacity of the water, and ΔT is the change in temperature in Celsius.
According to the problem:
The mass of the water (m) is 20 grams.
The initial temperature (T₀) is 30°C.
The specific heat of water (c) is 1 cal/g°C.
The heat absorbed (Q) is 300 calories.
To find ΔT, the change in temperature, we can re-arrange the formula to ΔT = Q / (mc). Plugging in the values we get: ΔT = 300 cal / (20 g × 1 cal/g°C) = 15°C.
The final temperature (Tₒ) is the initial temperature plus the change in temperature: Tₒ = T₀ + ΔT = 30°C + 15°C = 45°C.