Calculate the final temperature of 283 mL of water initially at 30 degrees celcius upon absorption of 19 kJ of heat
q = mass water x specific heat water x (Tfinal-Tinitial)
19,000 = 283 x 4.184 x (Tf - 30)
Solve for Tfinal.
31.6
To calculate the final temperature of the water, we can use the equation:
q = mcΔT
Where:
- q is the heat absorbed by the water,
- m is the mass of the water (which we can derive from the volume and density),
- c is the specific heat capacity of water, and
- ΔT is the change in temperature.
Let's break down the calculation step-by-step:
1. Calculate the mass of the water using the volume and density:
- The density of water is approximately 1 g/mL.
- Given that the volume is 283 mL, the mass can be calculated as follows:
Mass (m) = volume * density
= 283 mL * 1 g/mL
= 283 g
2. Determine the specific heat capacity of water:
- The specific heat capacity of water is approximately 4.18 J/g°C.
3. Plug the values into the formula and solve for the change in temperature (ΔT):
- q = mcΔT
- Rearrange the equation to solve for ΔT:
ΔT = q / (mc)
- Substitute the given values into the equation:
ΔT = 19,000 J / (283 g * 4.18 J/g°C)
4. Calculate ΔT:
- ΔT = 19,000 J / (118,894 J/°C)
- ΔT ≈ 0.16°C
5. Calculate the final temperature:
- The final temperature can be obtained by adding ΔT to the initial temperature:
Final temperature = 30°C + 0.16°C
≈ 30.16°C
Therefore, the final temperature of the water, upon absorbing 19 kJ of heat, is approximately 30.16 °C.