Question

The temperature of 1.00 mL of water is raised by 1.00°C for every 4.184 joules of heat absorbed by the water. How many liters of water can be raised from 21.0°C to 100.0°C by the absorption of 8.88 kcal of heat generated by the combustion of natural gas?

Answer 1

To raise the temperature of water from 21.0°C to 100.0°C using 8.88 kcal of heat, we can raise approximately 111.78 liters of water.

Step-by-step explanation:

To find out how many liters of water can be raised from 21.0°C to 100.0°C by the absorption of 8.88 kcal of heat, we need to convert the kcal to joules and then use the equation Q = mcΔT. First, let's convert the kcal to joules. Since 1 kcal = 4184 J, we can multiply 8.88 kcal by 4184 J/kcal to get 37063.52 J. Now we can use the equation Q = mcΔT, where Q is the heat absorbed, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature. We are given that ΔT = 100.0°C - 21.0°C = 79.0°C. Since the specific heat of water is 4.184 J/g°C, we can rearrange the equation to solve for the mass of water, m, which is equal to Q / (cΔT). Plugging in the values, we get m = 37063.52 J / (4.184 J/g°C * 79.0°C) = 111.78 g. Now we need to convert grams to liters. Since the density of water is 1 g/mL, 111.78 g is equal to 111.78 mL. Therefore, we can raise approximately 111.78 liters of water from 21.0°C to 100.0°C by the absorption of 8.88 kcal of heat.