Apr 16, 2015 · Hey ya'll my question reads: Use the formula Q = cmt to show that 300 cal are requited to raise the temperature of 300g of water from 22 degrees C to 30 degrees C. For the specific heat capacity c, use 1 cal/g degrees C. now i don't see how this math adds up, doesn't...

May 17, 2010 · Homework Statement How much heat (in kcal) must be removed to make ice at –10°C from 2 kg of water at 20°C? (The specific heat of ice is 0.5 cal/g °C.) Homework Equations Q=cmT The Attempt at a Solution I thought this problem was simple, Q=0.5x2000x30 Then diving by 1000 again...

Feb 26, 2015 · Homework Statement The water equivalent in kg, of a calorimeter having a mass of 0.3 kg and a specific heat of 0.2 is Homework Equations q=cmT The Attempt at a Solution I wanted to set cmT = cmT for the water and calorimeter, but I don't know what the equilibrium temperature is. So what do I do?

Nov 11, 2009 · a) Calculate q H20. I used the equation q = cmt, so q = (4.18)(53.0)(9.70) = 2150J b) Find the change of enthalpy for the reaction as it occurred in the calorimeter. = -2150J c) Find the change of enthalpy for the solution of 1.00g NaOH in water. For this I just halved -2150J and got -1080 (3 significant figures) but I'm not sure if this is ...

Feb 11, 2004 · I'm assuming that you would use the equation T=Q/cm and when I do, I end up with an answer of .95 degrees celcius which I know is wrong. Can anyone explain this to me? 2)When resting, a person has a metabolic rate of about 2.34 x 105 joules per hour. The person is submerged neck-deep into a tub containing 1.81 x 103 kg of water at 26.8 °C.

Nov 29, 2006 · Steam: Q=2.26*10^6 * m (m is what I'm trying to find) The amount of heat being released from the steam condensing to water is the amount of heat that raises the temperature of the water from 22C to 25C. Water: Q=cmT Q=(4178 J/kg*K)*(1.8 kg)*(25-22) Q=22561.2 J Then, I set the Q's equal to each other to get 0.009, which is incorrect.

May 24, 2007 · The latent heat of vaporization of H2O at body temperature (37.0°C) is 2.38E+6 J/kg. To cool the body of a 71.9 kg jogger [average specific heat capacity = 3550 J/(kg*°C)] by 1.10°C, how many kilograms of water in the form of sweat have …

Jul 18, 2009 · I'm not entirely sure either. But using Q=cmT to find out the energy needed to heat the water, and then dividing by the intercepted energy. Q=4x(1.08x10^26)x100 =4.32x10^26 Then divide by the intercepted power. Which converts to J/s. The answer comes out wrong though.

Dec 3, 2009 · I had originally tried to use the equation Q=cmT, with the values plugged in as Q=4186*1*150, giving me 627.9 kilocalories, but this is incorrect. Physics news on Phys.org Record-setting electron beam: Five times more powerful than predecessors

Dec 3, 2007 · Q=mcT where m is the density times the volume then with Q i found Power which is Q/t then with power i plugged it into the intensity equation of I=P/A (p is power not pressure and A is area) and solved for A, but for somereason I am not getting the right answer...is it a problem with conversion or is the process I am using wrong...