Final answer:
The heat associated with 110.0 g of water increasing in temperature from 24.5°C to 56.0°C is 14487.24 J. The specific heat of water (4.184 J/g°C) is used along with the mass of water and the temperature change to calculate this value.
Step-by-step explanation:
To calculate the heat (q) associated with the temperature increase of water, we need to use the formula q = m * c * ΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity (c) of water is commonly given as 4.184 J/g°C. First, we convert the mass of water from grams to kilograms by dividing it by 1000, yielding 0.110 kg. Next, we find the change in temperature (ΔT) by subtracting the initial temperature from the final temperature: 56.0°C - 24.5°C = 31.5°C. Finally, we multiply these values to get the heat in joules:
q = m * c * ΔT = (110.0 g) * (4.184 J/g°C) * (31.5°C)
q = 110.0 g * 4.184 J/g°C * 31.5°C = 14487.24 J
Since the water's temperature increased, it absorbed heat, and therefore, q is positive.