Video: Calculating the Energy Needed to Heat a Mass by a Known Temperature Change

Determine how much energy is needed to heat 2 kg of water by 3°C. Use a value of 4184 J/kg ⋅ °C for the specific heat capacity of water. Give your answer to 2 significant figures.

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Video Transcript

Determine how much energy is needed to heat two kilograms of water by three degrees Celsius. Use a value of 4184 joules per kilogram degrees Celsius for the specific heat capacity of water. Give your answer to two significant figures.

Okay, so in this exercise, we start out with two kilograms of water. And we want to heat this water up so its temperature raises by three degrees Celsius. So whatever the temperature is now, we wanna increase that by three degrees Celsius. We’ll call this change in temperature Δ𝑇. And we’ll label the mass of this water, two kilograms, as 𝑚. Given the specific heat capacity of water, we want to know how much energy is needed to heat this amount of water by this much.

There is a mathematical relationship we can recall to help us solve for that. That relationship tells us that the amount of energy needed to increase the temperature by some amount, Δ𝑇, of some amount of a substance, 𝑚, is equal to the product of those two values times the specific heat capacity of that substance. And we recall that, in general, specific heat capacity tells us how much energy is needed to heat up one kilogram of a given material by one degree Celsius. For a liquid water, that amount is 4184 joules.

In other words, if we add that number of joules of energy to one kilogram of water, then we’ll raise its temperature by one degree Celsius. So knowing the mass of our water, 𝑚, the change in temperature we want to affect, Δ𝑇, and the specific heat capacity of our water, we’re ready to calculate the energy needed. Plugging these values in for 𝑚, 𝐶, and Δ𝑇, let’s take a look at the units for a moment.

Notice that we have units of mass in kilograms in both numerator and denominator. That means those units will cancel out. We also have units of temperature in degrees Celsius in both denominator and numerator. So those units cancel as well. When we calculate this product, we’ll be left simply with units of joules, the units of energy. This confirms that we’re on the right track. And when we calculate this product, we find the result of 25104 joules.

This isn’t our final answer though because we see in the problem statement that we’re to give our answer to two significant figures. The first significant figure in this number is the two. The next significant figure is the five. And the third significant figure is the one. We’ll just keep two of these. And since our third significant figure is less than five, that means we won’t round up. We’ll keep our first two significant figures just as they are, two and five. Rounding our result to two significant figures, we see that it’s 25000 joules. That’s how much input energy would be needed to heat two kilograms of water by three degrees Celsius.

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