Calculate Energy Applied to Steel Slab

Thread starter debo22
Start date Aug 19, 2014
Tags

Applied Energy Thermal Thermal energy

AI Thread Summary

To calculate the energy applied to a slab of steel with a temperature rise described by the equation -13t + 5t^2 + 10*sin(3ωt), the energy can be determined using the formula mc(dT), where m is the mass, c is the specific heat, and dT is the change in temperature. The derivative of the temperature function, dT/dt, is found to be 13 + 10t + 30ω*cos(3ωt). The discussion raises the question of whether mass (m) and specific heat (c) are constant throughout the process. Clarification on the constancy of these parameters is essential for accurate energy calculations. Understanding these variables is crucial for applying the energy equation effectively.

Aug 19, 2014

debo22

I want to calculate the energy applied to a slab of steel with a temperature rise that varying with time (t) as -
13t + 5t^2 + 10* sin (3ωt)

The energy applied to the system is calculated by mc (dT) equation where
m mass of the slab
c specific heat
dT change in temperature

Science news on Phys.org

Aug 19, 2014

HallsofIvy

Science Advisor

Homework Helper

Are m and c constant?

Then dt= 13+ 10t+ 30\omega cos(3\omega t), of course.

Thread 'Condensate rate of water for an air conditioning cooling coil'

I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...

View full post »

Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'

Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...

View full post »