Question

an electric resistance carrying electric current 2A and potential difference 50V immersed in 200g of water durin 5 minutes
find
1-the change in termperature of water at the end of time
2-the rate of heat follow
3-the temperature rate

Answer 1

Question #42883 – Physics – Molecular Physics | Thermodynamics

Question.

An electric resistance carrying electric current 2A and potential difference 50V immersed in 200g of water during 5 minutes

find

1 - the change in temperature of water at the end of time

2 - the rate of heat flow

3 - the temperature rate

Given:

$$I = 2 \, \text{A}$$ $$U = 50 \, \text{V}$$ $$m = 200 \, \text{g} = 0.2 \, \text{kg}$$ $$t_{\text{end}} = 5 \, \text{min} = 300 \, \text{s}$$ $$c = 4200 \, \frac{\text{J}}{\text{kg} \cdot {}{}^{\circ}\text{C}}$$

Find:

1)

$$\Delta T (t_{\text{end}}) = ?$$

2)

$$\frac{Q}{t} = ?$$

3)

$$\frac{dT}{dt} = ?$$

Solution.

Work of the electric element is the amount of energy for the electric heating element by passing electrical current therethrough:

$$A = I U \Delta t$$

All work of electric element is used to heat water. Therefore:

$$A = Q$$ $$Q = m c \Delta T (t)$$ $$I U \Delta t = m c \Delta T (t)$$

So,

1)

$$I U t _ {e n d} = m c \Delta T (t _ {e n d}) \rightarrow \Delta T (t _ {e n d}) = \frac {I U t _ {e n d}}{m c}$$

Calculate:

$$\Delta T (t _ {e n d}) = \frac {2 \cdot 5 0 \cdot 3 0 0}{0 . 2 \cdot 4 2 0 0} = 3 5. 7 {}^ {\circ} \mathrm {C}$$

2) In our case, the rate of heat flow is the electrical power $N$ :

$$\frac {Q}{t} = N = I U$$

Calculate:

$$\frac {Q}{t} = 2 \cdot 5 0 = 1 0 0 W$$

3)

$$I U \Delta t = m c \Delta T$$ $$\frac {d T}{d t} = \frac {\Delta T}{\Delta t} = \frac {I U}{m c}$$

Calculate:

$$\frac {d T}{d t} = \frac {2 \cdot 5 0}{0 . 2 \cdot 4 2 0 0} = 0. 1 2 \frac {\circ \mathrm {C}}{s}$$

Answer.

1) the change in temperature of water at the end of time

$$\Delta T (t _ {e n d}) = \frac {I U t _ {e n d}}{m c} = 3 5. 7 {}^ {\circ} \mathrm {C}$$

2) the rate of heat flow

$$\frac {Q}{t} = I U = 1 0 0 W$$

3) the temperature rate

$$\frac {d T}{d t} = \frac {I U}{m c} = 0. 1 2 \frac {\circ \mathrm {C}}{s}$$

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