Question

For Water its temperature is 50°C and room temperature is 30°C just after 5 minutes excess temperature reduce to half its value,calculate temperature of water next 2.5 minutes

Answer 1

Answer on Question #44319-Physics-Molecular Physics-Thermodynamics

For water its temperature is $50{}^{\circ}\mathrm{C}$ and room temperature is $30{}^{\circ}\mathrm{C}$ just after 5 minutes excess temperature reduce to half its value, calculate temperature of water next 2.5 minutes

Solution

Let's use Fourier's law for heat flow:

$$\dot {q} = k (T _ {w} - T _ {r}),$$

where $T_{w}$ is the temperature of the water, $T_{r}$ is the room temperature, $k$ is the thermal conductivity coefficient.

Solving this equation gives:

$$T _ {w} - T _ {r} = A e ^ {- B t},$$

where $A$ and $B$ are constants.

When $t = 0$

$$T _ {w} - T _ {r} = A e ^ {- B \cdot 0} = A = 5 0 {}^ {\circ} C - 3 0 {}^ {\circ} C = 2 0 {}^ {\circ} C.$$

When $t = 5$ minutes

$$T _ {w} - T _ {r} = \frac {2 0 {}^ {\circ} C}{2} = 1 0 {}^ {\circ} C = 2 0 {}^ {\circ} C e ^ {- B \cdot 5 \min } \rightarrow B = \frac {\ln 2}{5} \min ^ {- 1}.$$

The temperature of water next 2.5 minutes is

$$T _ {w} (7. 5 \min) = 3 0 {}^ {\circ} \mathrm {C} + 2 0 {}^ {\circ} \mathrm {C} \cdot e ^ {- \frac {\ln 2}{5} 7. 5} = 3 0 {}^ {\circ} \mathrm {C} + 2 0 {}^ {\circ} \mathrm {C} \cdot (e ^ {- \ln 2}) ^ {1. 5} = 3 0 {}^ {\circ} \mathrm {C} + 2 0 {}^ {\circ} \mathrm {C} \cdot \left(\frac {1}{2}\right) ^ {1. 5} = 3 7 {}^ {\circ} \mathrm {C}.$$

Answer: $37{}^{\circ}C$

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