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1500-W heater is connected to a 120-V line for 2.0 hours. How much heat energy is produced?
A) 1.5 kJ B) 11 MJ C) 0.18 MJ D) 3.0 kJ
♦ Relevant knowledge
The formula below can be used to calculate the amount of energy or heat required to raise a material’s temperature without changing its phase taking into account the mass, the specific thermal capacity and the temperature differences.
Q=mcΔT
Where,
- The temperature variation is ΔT
- This mass is m and the c is the temperature constant that is specific to the particular.
This problem can be solved by understanding the relationship between power, heat used, and time that the heater works.
First, convert the hours into seconds to determine how the heater works.
Add the time to power and you will get the heat energy produced.
Here is the relationship between heat produced and power required by a heater:
[katex]Q = Pt[/katex]
Q refers to the heat produced by a heater, P refers to the power used by the heater and t indicates the time that the heater is on.
Here’s how to calculate the time it takes for the heater to work in seconds:
[katex]\begin{array}{c}\\t = \left( {2.0{\rm{ hours}}} \right)\left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ hour}}}}} \right)\\\\ = 7200{\rm{ s}}\\\end{array}[/katex]
The heater can be used for up to 7200 s.
The following equations can be used to calculate the heat energy generated by the heater:
[katex]Q = Pt[/katex]
In the equation above, substitute 1500 W for P and 7200 S for t.
[katex]\begin{array}{c}\\Q = \left( {1500{\rm{ W}}} \right)\left( {7200{\rm{ s}}} \right)\left( {\frac{{{\rm{J}} \cdot {{\rm{s}}^{ – 1}}}}{{1{\rm{ W}}}}} \right)\\\\ = \left( {10.8 \times {{10}^6}{\rm{ J}}} \right)\left( {\frac{{{{10}^{ – 6}}{\rm{ MJ}}}}{{1{\rm{ J}}}}} \right)\\\\ = 10.8{\rm{ MJ}}\\\\ \approx 11{\rm{ MJ}}\\\end{array}[/katex]
Correct answer is therefore option-B) 11 MJ.
Ans:
Option-B) 11 MJ is the correct answer.