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Question

How much heat is needed to raise the temperature of 122.5 g of water by 14.0°C? Please provide your final answer rounded to two decimal places and provide the proper units, in their respective boxes(Hint: You have the specific heat (cof water in your notes!

Answer 1

Hello, Camila,

We need the specific heat of water. The value we'll use here is 4.184 J/gK. Other values are available in different units of heat, mass and temperature, but the are all equivalent once the conversions are taken into account. For example, they can be found with moles instead of grams, calories in place of Joules, and C instead of K. You can use any of them, but be sure do convert the raw data into units that are compatible.

Specific heats are unique to materials, and can sometimes be used to help identify a substance. The value of 4.184 J/gK means that it takes 4.184 Joules of energy to heat 1 gram of water by 1 degree K. Iron (Fe), by contrast, has a specific heat of only 0.45 J/gK. That sounds counterintuitive to most, so if it makes sense to you, Congratulations. Many might predict should have a high specific heat. The nice thing about metals and their low specific heats is that the heat doesn't linger - it is transmitted through the pan or skillet quickly to the food and water for cooking. The metal pan wants to send the heat through quickly. Water, on the other hand, takes forever and a day to reach boiling, much to the relief of my noodles.

The equation we can use for answering this problem is:

q = mCDT

where m is the mass (or moles, depending on the units of C), C is the specific heat, in J/gK here, and DT is the Delta Temperature - the temperature change (T₂ - T₁).

Enter the data into the equation and see if the units cancel to give what we want, Joules, the energy required to raise the mass of water by 14.0C.

q = (122.5g)*(4.184J/gK)*(14K)

[Note: the conversion of C to K simply adds 273 to both the initial and final temperatures, so the difference of 14.0 is the same]

Cancel units and solve:

q = (122.5g)*(4.184J/gK)*(14K)

q = 7176 Joules (but we are limited to 3 sig figs, so 7180 Joules is correct.)

Bob

Note that this can be done in reverse. The question may have been how much energy needs to be removed to lower the temperature by 14.0K. In this case the DT is negative (T₂ - T₁), so the energy would also be negative (-7176 Joules). The minus sign means that energy is being REMOVED for the system (i.e., it gets colder.

Answer 2

I will assume you want the result in joules. The equation necessary to solve this problem is Q = c x m x Δt where c is the specific heat, m the mass, and Δt the temperature change in C. I looked up the specific heat of water in joules and found it to be 4184 j/kg-C. We can change this ot 4.184 j/g-2 to suit this problem.

Q = 4.184 j/C-g x 122.5 g x 14.0 C = 7176 joules.