Question

A 1.5 kW hotplate is used to heat 250ml of water in a 292.1g glass beaker both the water and the glass beaker are initially at 20.5C the hotplate is used for 5.5 minutes and produces 495kJ of energy when used. Determine the final temperature of the glass beaker

Answer 1

Answer on Question #49798, Physics, Molecular Physics

A 1.5 kW hotplate is used to heat 250ml of water in a 292.1g glass beaker both the water and the glass beaker are initially at 20.5C the hotplate is used for 5.5 minutes and produces 495kJ of energy when used. Determine the final temperature of the glass beaker.

Solution

According to the law of conservation of energy, the energy “produced” by hotplate was used to heat water and glass beaker.

$$Q = Q_W + Q_G,$$

where $Q$ is the energy “produced” by hotplate; $Q_W$ is the energy used to heat water; $Q_G$ is the energy used to heat the glass beaker.

If the water is heated to 100 °C, it starts boiling. We find the amount of heat required to heat the water from 20.5 to 100 °C (see Eq. (2))

$$Q_W = c_W m_W (t - t_0),$$

where $t$ the final temperature of water; $t_0$ the initial temperature of water; $c_W$ is the specific heat of water ($c_W = 4200\,J / (kg \cdot C)$); $m_W$ is the mass of water ($m_W = V\rho$); $\rho$ is the density of water ($\rho = 1000\,kg / m^3$); $V$ is the volume of water ($250\,ml = 0.250l = 2.5 \cdot 10^{-4}\,m^3$).

$$Q_W = c_W m_W (t - t_0) = 4200\,J / (kg \cdot C) \cdot 1000\,kg / m^3 \cdot 2.5 \cdot 10^{-4}\,m^3 (100 - 20.5) = 83475\,J = 83.475\,kJ$$ $$Q_G = c_G m_G (t - t_0),$$

where $t$ the final temperature of the glass beaker; $t_0$ the initial temperature of the glass beaker; $c_G$ is the specific heat of the glass ($c_G = 840\,J / (kg \cdot C)$).

$$Q_G = c_G m_G (t - t_0) = 840\,J / (kg \cdot C) \cdot 0.2921\,kg \cdot (100 - 20.5) = 19506.4\,J = 19.5064\,kJ$$

It is clear that $Q = Q_W + Q_G$, $Q_W + Q_G = 102.981\,kJ$, $Q = 495\,kJ$

The water will be converted into steam. The heat required to convert 250ml of water entirely in steam (see Eq. (4)).

$$Q_S = m_W L$$

$L$ is the specific heat of vaporization ($L = 2256\,kJ / kg$).

$$Q _ {S} = m _ {W} L = 1000 \mathrm{kg} / \mathrm{m} ^ {3} \cdot 2.5 \cdot 10 ^ {- 4} \mathrm{m} ^ {3} \cdot 2256 \mathrm{kJ} / \mathrm{kg} = 564 \mathrm{kJ}$$

It is clear that $Q_{S} > \left(Q - \left(Q_{W} + Q_{G}\right)\right)$.

Not all water is transformed into steam. During the evaporation temperature of the system will remain unchanged (100 ° C)

Answer: the temperature of the glass beaker will be 100°C

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