Question

An electrical heater is used to heat 100g of water in a well-insulated container at a steady rate. The temperature of the water increases by 15 degrees C when the heater is operated for a period of 5.0 minutes. Determine the change of temperature of the water when the same heater and container are individually used to heat 100g of water for a time of 2.5 minutes.

Answer 1

The key here is "steady rate." This indicates a linear slope between the change in the temperature of the water and time.

Let's set up a linear equation\[\Delta T = {15 \over 5} t=3 t\]

Plug in 2.5 and you should get your answer. (It should be half the delta T when heated for 5 minutes.)

Answer 2

This is the wrong answer. You have'nt taken the mass of water into account!

Answer 3

We cannot account for the mass of the water directly without knowing the input electrical energy into the heater and efficiency of the heater, or the output energy of the heater \[Q = E_{elec} \eta_{heater} = m_{w} c_{w} \Delta T_w\]

Let's say we knew the power output of the electric heater. \[P = {Q \over t}\]Then we could set up the following expression, \[P t = m_w c_w \Delta T_w\]We can find the power output of the electric heater from the given values for the 5 minutes case. \[P = {{0.1} \cdot 4.184 \cdot 15 \over 5}\]
Let's substitute this back in for the 2.5 minutes case. \[{0.1 \cdot 4.184 \cdot 15 \over 5} \cdot 2.5 = 0.1 \cdot 4.184 \cdot \Delta T\]Realizing that mass and specific heat cancel out\[{2.5 \over 5} \cdot 15 = \Delta T\]\[{1 \over 2} \cdot 15 = \Delta T\]

This agrees with my earlier statement that the change in temperature should be half when ran for half the time.