2.5 kg of liquid water initially at 18°C is to be heated to 96°C in a teapot equipped with a 1200-W electric heating element inside. The teapot is 0.8 kg and has an average specific heat of 0.6 kJ/kg · °C. Taking the specific heat of water to be 4.18 kJ/kg · °C and disregarding any heat loss from

Question

2.5 kg of liquid water initially at 18°C is to be heated to 96°C in a teapot equipped with a 1200-W electric heating element inside. The teapot is 0.8 kg and has an average specific heat of [latex] 0.6 kJ / kg \cdot{ }^{\circ} C [/latex]. Taking the specific heat of water to be [latex] 4.18 kJ / kg \cdot{ }^{\circ} C [/latex] and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated.

Accepted Answer

Cold water is to be heated in a 1200-W teapot. The time needed to heat the water is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Thermal properties of the teapot and the water are constant. 3 Heat loss from the teapot is negligible.

Properties The average specific heats are given to be [latex] 0.6 kJ / kg \cdot{ }^{\circ} C [/latex] for the teapot and [latex] 4.18 kJ / kg \cdot{ }^{\circ} C [/latex] for water.

Analysis We take the teapot and the water in it as our system that is a closed system (fixed mass). The energy balance in this case can be expressed as

[latex] E_{ ext {in }}-E_{ ext {out }}=\Delta E_{ ext {system }} [/latex]

[latex] E_{ in }=\Delta U_{ ext {system }}=\Delta U_{ ext {water }}+\Delta U_{ ext {tea pot }} [/latex]

Then the amount of energy needed to raise the temperature of water and the teapot from 18°C to 96°C is

[latex] E_{ ext {in }}=(m C \Delta T)_{ ext {water }}+(m C \Delta T)_{ ext {teapot }} [/latex]

[latex] = (2.5 kg )\left(4.18 kJ / kg \cdot{ }^{\circ} C ight)(96-18){ }^{\circ} C +(0.8 kg )\left(0.6 kJ / kg \cdot{ }^{\circ} C ight)(96-18){ }^{\circ} C [/latex]

[latex] = 853 kJ [/latex]

The 1500 W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 853 kJ of heat is determined from

[latex] \Delta t=\frac{ ext { Total energy transferred }}{ ext { Rate of energy transfer }}=\frac{E_{ ext {in }}}{\dot{E}_{ ext {transfer }}}=\frac{853 kJ }{1.2 kJ / s } = 710 s = 11.8 min [/latex]

Discussion In reality, it will take longer to accomplish this heating process since some heat loss is inevitable during the heating process

Question 1.121: 2.5 kg of liquid water initially at 18°C is to be heated to ...

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