Question

Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr = 8 gm, and cr = 3.7 J/gm ∘K-1. If the water is initially at room temperature, how long will it take for the water to heat up 5∘K? (Hint: dT/dt is approximately equal to Δ T / Δt .)

Answer 1

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

$\frac{E}{t} = VI$

solving for t

$t = \frac{E}{VI}$

$t = \frac{m_w C_w \Delta T}{VI}$

$t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}$

t = 444.125 sec

Answer 2

Answer:

$t=5057.9167\ s$

Explanation:

Given:

• Voltage supply to the resistor, $V=24\ V$

• current supply to the resistor, $I=0.1\ A$

• mass of water, $m_w= 51\ g$

• specific heat of water, $c_w=4180\ J.kg^{-1}.K^{-1}$

• specific heat of resistor, $c_r=3700\ J.kg^{-1}.K^{-1}$

• mass of resistor, $m_r=0.008\ kg$

• change in temperature, $\Delta T=50\ K$

Now the amount of heat required to heat the water by 50 K:

$Q_w=m_w.c_w.\Delta T$

$Q_w=0.051\times 4180\times 50$

$Q_w=10659\ J$

Now the amount of heat required to heat the resistor by 50 K:

$Q_r=m_r.c_r.\Delta T$

$Q_r=0.008\times 3700\times 50$

$Q_r=1480\ J$

Now the total heat to converted from the electrical energy:

$Q=Q_w+Q_r$

$Q=12139\ J$

Now Using Joule's law of heating:

$Q=V.I.t$

$12139=24\times 0.1\times t$

$t=5057.9167\ s$