Question

People often confuse the concepts of "temperature" and "heat" This exercise is designed to illustrate that it is possible to be adding "heaf" to a system while its temperature remains constant A beaker contains 50 g of water initially at a temperature of 72.4°C. After 5 min, the temperature has fallen to 55.7°C. How much energy did the water lose? Give the answer in J. 1. What is the magnitude of the average rate of energy loss over the 5 minute interval? (This is power, since it is a rate of change of energy.) 2. At what average rate must energy be added to the water to keep the temperature constant? Explain. 3. 4. If we use a burner or some other device to put energy into the water at the rate specified in 3, is heat being added to the water? changing? Is the temperature

Answer 1

Given data m=50 gm

Here we know 1 gram=0.001 kg

Here, 50 grams are given so,

i.e.,50*0.001=0.05 kg

T ₁=72.4°

T ₂=55.7°

t=5 min

we know 1 min=60 sec

t=5*60=300 sec

S=4186 J/kg/k

(1) Finding how much energy did water lose

We know E=ms(T ₁-T ₂)

                     =0.05*4186(72.4-55.7)

                    =209.3(16.7)

E=3495.31 J

                   

(2) calculating the magnitude of average rate of energy loss over the 5 minute interval

                 P=E/t

                   =3495.3/300

P=11.65 w

(3) average rate must energy to be added to water

Average rate=11.65 w to be added

(4) Yes, if we use a burner or some other device to put energy into the water heat is added

NO, temperature will not change