Answer to Question #191436 in Classical Mechanics for Kaitlyn Hosking

Question

1. Calculate the amount of energy required to raise the temperature of
2. 0.5kg of water from 20⁰C to 65⁰C.  (Specific heat capacity of water = 4180J/kg⁰C)
3. A 950g steel baking tray from 18⁰C to 183⁰C.  (Specific heat capacity of steel = 500J/kg⁰C)

1. A well insulated kettle contains 1.2kg of water at a temperature of 20⁰C.  The kettle is switched on and takes 180s to come to the boil. (Specific heat capacity of water = 4180J/kg⁰C)
2. How much energy is absorbed by the water in 180s?
3. Calculate the power rating of the kettle.
4. A cheaper kettle is not so well insulated, what effect will this have on the time to heat the water?

1. The tip of the soldering iron is made of copper with a mass of 30 g.  Calculate how much heat energy is required to heat up the tip of a soldering iron by 400 °C.

(specific heat capacity of copper = 380 J/kg°C)

Answer 1

1) 2.

$$Q=mc\Delta T=0.5\ kg\cdot4180\ Jkg^{-1}\ \!^{\circ}C^{-1}\cdot45^{\circ}C=94050\ J.$$

3.

$$Q=mc\Delta T=0.95\ kg\cdot500\ Jkg^{-1}\ \!^{\circ}C^{-1}\cdot165^{\circ}C=78375\ J.$$

1) 2.

$$Q=mc\Delta T=1.2\ kg\cdot4180\ Jkg^{-1}\ \!^{\circ}C^{-1}\cdot80^{\circ}C=401280\ J.$$

3.

$$Q=Pt,$$ $$P=\dfrac{Q}{t}=\dfrac{401280\ J}{180\ s}=2229\ W=2.2\ kW.$$

4. Since the cheaper kettle is not so well insulated, there will be the heat losses, therefore, it takes more time to heat the water.

1)

$$Q=mc\Delta T=0.03\ kg\cdot380\ Jkg^{-1}\ \!^{\circ}C^{-1}\cdot400^{\circ}C=4560\ J.$$