A total of 506 J of heat are added to 50.0 g of water initially at 15.0 degrees Celsius. What is the final temperature of the water?
506=50*c*(tf-15)
find the specific heat of water, in joules/g-C. Actually, you ought to have it memorized.
To find the final temperature of the water, we can use the equation:
q = m * C * ΔT
Where:
- q is the heat energy absorbed or released by the substance (in joules),
- m is the mass of the substance (in grams),
- C is the specific heat capacity of the substance (in J/g°C),
- ΔT is the change in temperature of the substance (in °C).
In this case, the water is absorbing heat energy (q = +506 J), the mass is 50.0 g, and the initial temperature is 15.0°C.
First, we need to find the specific heat capacity of water, which is 4.18 J/g°C.
Rearrange the equation to solve for ΔT:
ΔT = q / (m * C)
Now, substitute the given values into the equation:
ΔT = 506 J / (50.0 g * 4.18 J/g°C)
Simplify the equation:
ΔT = 2.42°C
Finally, to find the final temperature, add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 15.0°C + 2.42°C
Therefore, the final temperature of the water is 17.42°C.