Question

1. How long would you expect a 2.5 kW heater to take to heat 0.6 L of water from 20 to 100°C if there were no heat losses.

2. How much water will boil away in the last 20 seconds of the heating?

Answer 1

Answer on Question#38678, Physics, Molecular Physics

1. Heat, obtained from heater in time $t$ is $Q = P \cdot t$ by definition of power. From the other side, amount of heat needed to rise temperature of water from 20 to 100 degrees is

$Q = cm\Delta T = c\rho_{\mathrm{w}}V\Delta T$ , where $c$ is heat capacity, $\rho_{\mathrm{w}}$ is the density of the water, $V$ is volume of water and $\Delta T = T_1 - T_2$ is the difference between initial and final temperatures. Using last two equations, obtain:

$$P t = c \rho_ {\mathrm {w}} V \Delta T \Rightarrow t = \frac {c \rho_ {\mathrm {w}} V \Delta T}{P} = \frac {4 2 0 0 \frac {J}{k g \cdot K} \cdot 1 0 0 0 \frac {k g}{m ^ {3}} \cdot 0 . 6 \cdot 1 0 ^ {- 3} m ^ {3} \cdot 8 0 K}{2 5 0 0 W} = 8 1 s.$$

2. Since in last 20 seconds water is not at 100 degrees, none of it will boil in last 20 seconds.