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A container holds 0.2 \mathrm{~kg} of water initially at a temperature of 25^{\circ} \mathrm{C}. Heat is added to the water until its temperature reaches 100^{\circ} \mathrm{C}. Calculate the heat added to the water during this process.
Given the specific heat capacity of water is 4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}.
 

Option: 1

62895 \mathrm{~J}


Option: 2

8170 \mathrm{~J}


Option: 3

9200 \mathrm{~J}


Option: 4

7820 \mathrm{~J}


Answers (1)

best_answer

The amount of heat (Q)added to a substance can be calculated using the formula:
\mathrm{Q=m c \Delta T}
where mis the mass of the substance, cis its specific heat capacity, and  \mathrm{\Delta T}is the change in temperature.

Given that \mathrm{ m=0.2 \mathrm{~kg}, c=4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K},} and the temperature change \mathrm{\Delta T=100^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}=75^{\circ} \mathrm{C},} we can calculate the heat added:
\mathrm{Q=0.2 \mathrm{~kg} \times 4186 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \times 75^{\circ} \mathrm{C}=62895 \mathrm{~J} }

Therefore, the heat added to the water during this process is \mathrm{62895 \mathrm{~J}}.

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Gaurav

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