Question

1. A 100kw electric heater heats up 100L of water at 20°C for 2 minutes. If 1L of water has a mass of 1kg find the final temperature of water. Specific heat capacity of water is 4200J/kg/°C.

2. A store hotplate is rated at 1kw. How long will it take for 1.5L (1.5kg) of water initially at 10°C to start to boil. Specific heat capacity = 4200J/kgk

Answer 1

Answer on Question 79559, Physics, Other

Question:

1) A $100\,kW$ electric heater heats up $100\,L$ of water at $20{}^{\circ}\mathrm{C}$ for 2 minutes. If $1\,L$ of water has a mass of $1\,kg$ find the final temperature of water. Specific heat capacity of water is $4200\,J/kg\cdot{}^{\circ}\mathrm{C}$.

Solution:

Let's first find the quantity of heat that the water needs in order to reach the final temperature during 2 minutes:

$$Q = mc\Delta t = mc(T_{final} - T_{initial}),$$

here, $m = 100\,kg$ is the mass of the water, $c = 4200\,J/kg\cdot{}^{\circ}\mathrm{C}$ is the specific heat capacity of the water, $T_{initial} = 20{}^{\circ}\mathrm{C}$ is the initial temperature of the water and $T_{final}$ is the final temperature of the water.

From the other hand, we can write:

$$Q = Pt,$$

here, $Q$ is the quantity of heat that the water needs in order to reach the final temperature during time $t = 2\,min$ and $P = 100\,kW$ is the power of the electric heater.

Finally, we can equate both expressions and find the final temperature of the water:

$$mc(T_{final} - T_{initial}) = Pt,$$ $$T_{final} = \frac{Pt}{mc} + T_{initial} = \frac{10^5\,W \cdot 2 \cdot 60\,s}{100\,kg \cdot 4200\, \frac{J}{kg \cdot {}{}^{\circ}\mathrm{C}}} + 20{}^{\circ}\mathrm{C} = 48.6{}^{\circ}\mathrm{C}.$$

Answer:

$$T_{final} = 48.6{}^{\circ}\mathrm{C}.$$

2) A store hotplate is rated at $1\,kW$. How long will it take for $1.5\,L$ (1.5 kg) of water initially at $10{}^{\circ}\mathrm{C}$ to start to boil. Specific heat capacity of water is $4200\,J/kg\cdot{}^{\circ}\mathrm{C}$.

Solution:

We can find the quantity of heat that the water needs to start to boil from the formula:

$$Q = mc\Delta t = mc(T_{final} - T_{initial}),$$

here, $m = 1.5 \, kg$ is the mass of the water, $c = 4200 \, J/kg \cdot {}{}^{\circ}\mathrm{C}$ is the specific heat capacity of the water, $T_{initial} = 10{}^{\circ}\mathrm{C}$ is the initial temperature of the water and $T_{final} = 100{}^{\circ}\mathrm{C}$ is the final temperature of the water.

From the other hand, we can write:

$$Q = P t,$$

here, $Q$ is the quantity of heat that the water needs to start to boil, $P = 1 \, kW$ is the power of the store hotplate and $t$ is the time that the water needs to start to boil.

Finally, we can equate both expressions and find the time that the water needs to start to boil:

$$\begin{array}{l} m c \left(T _ {f i n a l} - T _ {i n i t i a l}\right) = P t, \\ t = \frac {m c \left(T _ {\text {f i n a l}} - T _ {\text {i n i t i a l}}\right)}{P} = \frac {1 . 5 \, k g \cdot 4 2 0 0 \, \frac {J}{k g \cdot {} {}^ {\circ} \mathrm {C}} \cdot (1 0 0 {}^ {\circ} \mathrm {C} - 1 0 {}^ {\circ} \mathrm {C})}{1 0 ^ {3} \, W} = 5 6 7 \, s = \\ \end{array}$$

**Answer:**

$$t = 5 6 7 \, s = 9. 4 5 \, \text{min}.$$

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