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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

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
and the*m*is the temperature constant that is specific to the particular.*c*

Concepts and ReasonThis 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.

FundamentalsHere is the relationship between heat produced and power required by a heater:

[katex]Q = Pt[/katex]

Qrefers to the heat produced by a heater,Prefers to the power used by the heater andtindicates 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

Pand 7200 S fort.[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.