If 335 g of water at 65.5°C loses 9750 J heat, what is the final temperature of the water?

Question

Noah Aldrich · Accepted Answer

The final temperature is #=58.6ºC#

Let the final temperature be #=TºC#

Then,

#DeltaT=(65.5-T)ºC#

Mass of water, #m=0.335kg#

Heat lost, #H=9.750kJ#

specific heat of water, #s=4.186kJkg^-1ºC^-1#

Therefore,

#H=msDeltaT#

#9.750=0.335*4.186*(65.5-T)#

#65.5-T=9.750/(0.335*4.186)=6.95#

#T=65.5-6.95=58.6ºC#

Ellie Andrews · Answer

undefined The final temperature of the water is 57.1°C.

Xavier Avery · Answer

undefined To find the final temperature of the water, you can use the formula for heat transfer:

$$ Q = mc\Delta T $$

Where:
- $ Q $ is the heat transferred (9750 J in this case)
- $ m $ is the mass of the water (335 g)
- $ c $ is the specific heat capacity of water (4.18 J/g°C, approximately)
- $ \Delta T $ is the change in temperature (final temperature minus initial temperature)

We can rearrange the formula to solve for $ \Delta T $:

$$ \Delta T = \frac{Q}{mc} $$

Substitute the given values:

$$ \Delta T = \frac{9750 \, \text{J}}{335 \, \text{g} \times 4.18 \, \text{J/g°C}} $$

$$ \Delta T \approx 6.68°C $$

To find the final temperature, subtract $ \Delta T $ from the initial temperature:

$$ \text{Final temperature} = 65.5°C - 6.68°C $$

$$ \text{Final temperature} \approx 58.82°C $$

Therefore, the final temperature of the water is approximately 58.82°C after losing 9750 J of heat.