Question

What is the minimum amount of heat (in joules) needed to increase the temperature of 6 kg of liquid water from 33 degrees Celsius to 71 degrees Celsius?

A) 1,656,000 J
B) 2,118,000 J
C) 1,254,000 J
D) 1,890,000 J

Answer 1

The minimum amount of heat needed is approximately 1,588,000 joules.

Step-by-step explanation:

To calculate the minimum amount of heat needed to increase the temperature of water, we can use the equation:

Q = mc∆T

where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ∆T is the change in temperature.

Given that the mass of the water is 6 kg, the specific heat capacity of water (c) is 4.184 J/g °C, and the change in temperature (∆T) is 71 - 33 = 38 degrees Celsius, we can substitute these values into the equation to find the minimum amount of heat:

Q = (6 kg)(1000 g/kg)(4.184 J/g °C)(38 °C) = 1,587,888 J ≈ 1,588,000 J

Therefore, the minimum amount of heat needed to increase the temperature of 6 kg of liquid water from 33 degrees Celsius to 71 degrees Celsius is approximately 1,588,000 joules.