Question
A stream of warm water is produced in a steady-flow mixing process by combining $1.0 \mathrm{kg} \mathrm{s}^{-1}$ of cool water at $298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)$ with $0.8 \mathrm{kg} \mathrm{s}^{-1}$ of hot water at $348.15 \mathrm{K}$$\left(75^{\circ} \mathrm{C}\right) .$ During mixing, heat is lost to the surroundings at the rate of $30 \mathrm{kW}$. What is the temperature of the warm-water stream? Assume the specific heat of water constant at $4.18 \mathrm{kJ} \mathrm{kg}^{\prime} \mathrm{K}^{-1}$.
Step 1
The mass flow rate of the warm water is the sum of the mass flow rates of the hot and cold water. So, we have: \[m_3 = m_1 + m_2 = 1.0 \, \mathrm{kg/s} + 0.8 \, \mathrm{kg/s} = 1.8 \, \mathrm{kg/s}\] Show more…
Show all steps
Your feedback will help us improve your experience
Supratim Pal and 84 other Chemistry 102 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
15 kg of water is heated from 67°C to 100°C. What is the change in entropy? The specific heat of water is 10 cal/kg·K. Answer in units of cal/K.
Water at a temperature of 20°C flows from a tap T into a container C at a rate of 0.021 liters per minute. The heating element (resistor R) generates sufficient power P so that the water in the container boils and gets converted into steam. At steady state, it is found that the amount of water in the container neither increases nor decreases with time. Neglecting heat losses, the value of P (in W) is given by (to the nearest integer) [For water specific heat capacity = 4.2 kJ/kg K; latent heat of vaporization = 2.3 MJ/kg]
If $125.6 \mathrm{kJ}$ of heat are supplied to $5.00 \times 10^{2} \mathrm{g}$ of water at $22^{\circ} \mathrm{C},$ what is the final temperature of the water?
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD