What mass of water will change its temperature by 3 C when 525 J of heat is added to it? The specific heat of water is 4-.8 J/gC

Question

Wyatt Atkins · Accepted Answer

#m= 36.5g#

To obtain the mass of water, let's use the equation below:

Based on the information you've provided, we know the following variables:

#DeltaT# = #3^oC#
#Q# = 525 J
#C# = #4.8J/(gxx^oC)#

All we have to do is rearrange the equation to solve for m. This can be accomplished by dividing both sides by #C# and #Delta T# to get m by itself like this:

#m = Q/(DeltaTxxC)#

Now, we just plug in the known values:

#m = (525cancelJ)/(3^cancel"oC"xx4.8cancelJ/(gxx^(cancel"oC")))#

#m = 36.5 g #

Serenity Allen · Answer

undefined To find the mass of water that will change its temperature by 3°C when 525 J of heat is added, you use the formula:

$$q = mcΔT$$

Where:
- $q$ = heat energy (in joules)
- $m$ = mass of the substance (in grams)
- $c$ = specific heat capacity of the substance (in J/g°C)
- ΔT = change in temperature (in °C)

Rearrange the formula to solve for mass:

$$m = \frac{q}{cΔT}$$

Substitute the given values:

$$m = \frac{525 \, \text{J}}{(4.18 \, \text{J/g°C}) \times 3 \, \text{°C}}$$

$$m \approx 40 \, \text{grams}$$

So, the mass of water that will change its temperature by 3°C when 525 J of heat is added is approximately 40 grams.