# Video: Calculating the Energy Required to Raise the Temperature of a Given Mass of Water, Given the Initial and Final Temperatures

How much energy is required to raise the temperature of 25 g of water from 280 K to 330 K? [A] 1,510 J [B] 2,590 J [C] 3,550 J [D] 4,180 J [E] 5,230 J

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### Video Transcript

How much energy is required to raise the temperature of 25 grams of water from 280 kelvin to 330 kelvin? A) 1,510 joules, B) 2,590 joules, C) 3,550 joules, D) 4,180 joules, or E) 5,230 joules.

In this question, we need to figure out how much energy it will take to raise the temperature of water. Different substances require different amounts of energy to raise their temperature. To find the amount of energy that’s needed to raise the temperature of a substance. We can use this formula which says that the energy that’s needed to raise the temperature is equal to the substance’s specific heat capacity times the mass of the substance times the change in temperature.

The specific heat capacity of water is 4.18 joules per gram per degree Celsius. So, let’s plug everything in to solve for the energy. The specific heat capacity of water is 4.18 as we discussed. The mass of the water is 25 grams. So, now we just need to figure out the change in temperature. The specific heat capacity is given in units of degrees Celsius, but our temperatures are given in units of kelvin. Temperatures in kelvin are equal to the temperature in Celsius plus 273.

So, we can either convert these temperatures to degrees Celsius. Or we can simply recognize that the change in temperature in both kelvin and Celsius will be the same since to convert to kelvin, we’re just adding a number. So, the change in temperature will be 330 minus 280, which gives us 50 kelvin, which will be equal to the change in temperature in Celsius. So, now we can plug that in and solve for the energy. Both the units of grams and degrees Celsius cancel here, leaving us with units of joules which are the correct units for energy.

Doing the math here might be a little bit hard to do in our heads, but the answer choices are pretty spread out, so we can do a little rounding. Let’s round the specific heat capacity down from 4.18 to four. Four times 25 gives us 100. And 100 times 50 gives us 5,000. Since we rounded the specific heat capacity down to make the mental math easier, this answer will be lower than the real answer. So, this matches answer choice E. It requires 5,230 joules to raise the temperature of 25 grams of water from 280 kelvin to 330 kelvin.