Question

A system consisting of 57.5 g of liquid water at 298 K is heated using an immersion heater at a constant pressure of 1.00 bar. If a current of 1.50 A passes through the 10.0-ohm resistor for 150 s, what is the final temperature of the water

Answer 1

Answer:

Explanation:

Heat generated = I² Rt , I is current , R is resistance , t is time .

= 1.5 x 1.5 x 10 x 150

Q = 3375 J

Q = m s ΔT , m is mass , s is specific heat of water ,  ΔT is rise in temperature.

ΔT = Q /  m s

= 3375 / 57.5 x 4.2

= 14 k

Final temperature

= 298 + 14

= 312 k

Answer 2

Answer:

Final temperature =311.97 K

Explanation

Let us first calculate the energy released when current is passed

by using the formula

$E=I^2Rt$

where I =1.50 A , R = 10.0 Ω , t = 150 s

Now lets put these values in the above equation to get

$E=1.5^2\times10\times150$

=3375 J

To calculate the change in temperature

 Heat released( q ) = m× s × Δ T

m= mass of liquid =57.5 g, s= specific heat capacity of water = 4.2 J/g K

ΔT= difference in temperature =(T-298) K, considering T to be the final temperature.

therefore,

3375=57.5×4.2×(T-298)

⇒T= (3375+57.5×4.2×298)/57.5×4.2

T=311.97 K