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# Calculate the amount of heat energy required to raise the temperature of $100{\text{g}}$ of copper from $20^\circ {\text{C}}$ to $70^\circ {\text{C}}$. Specific heat capacity of copper $= 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$.A) $1950{\text{J}}$B) $3900{\text{J}}$C) $390{\text{J}}$D) ${\text{None of the above}}$

Last updated date: 17th Jun 2024
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Hint: An increase in temperature requires heat to be supplied. Now the amount of this heat will be proportional to the mass of the copper and the temperature difference between the current temperature of the copper and the temperature we desire. The specific heat capacity of copper will serve as the proportionality constant.

Formula used:
The amount of heat required is given by, $Q = mc\Delta T$ where $m$ is the mass of the sample, $c$ is the specific heat capacity of the material of the sample and $\Delta T$ is the change in temperature.

The mass of the sample of copper is given to be $m = 100{\text{g}}$ .
The specific heat capacity of copper is given to be $c = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
Since the temperature has to be raised from $20^\circ {\text{C}}$ to $70^\circ {\text{C}}$ , the change in temperature will be $\Delta T = 70 - 20 = 50^\circ {\text{C}}$ .
The amount of heat required to bring about the necessary change in temperature can be expressed as $Q = mc\Delta T$ --------- (1)
Substituting for $m = 0.1{\text{kg}}$ , $c = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ and $\Delta T = 50{\text{K}}$ in equation (1) we get, $Q = 0.1 \times 390 \times 50 = 1950{\text{J}}$
$\therefore$ the heat required to bring about the necessary change in temperature is obtained to be $Q = 1950{\text{J}}$ .
Note: While substituting values of different physical quantities in any equation make sure that all the quantities are expressed in their respective S.I units. If not, then the necessary conversion of units must be done. Here the mass of the sample was expressed in the unit of grams so we expressed it in the S.I unit of kilogram as $m = 0.1{\text{kg}}$ before substituting in equation (1). However, the change in the temperature is basically a difference between two temperatures and its value in the Celsius scale and the Kelvin scale are the same, so we do not make a conversion of units for the change in temperature $\Delta T$ .