Question

How much heat energy is required to raise the temperature of 0.368kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of copper is 0.0920 cal/(g⋅∘C)

Answer 1

The heat required to raise the temperature of the copper is $\boxed{1252.6\,{\text{cal}}}$.

Further Explanation:

Given:

The mass of the copper is $0.368\,{\text{kg}}$.

The specific heat capacity of copper is $0.0920\,\text{cal/g}\cdot^\circ\text{C}$ .

Concept:

The specific heat capacity of a particular substance is the amount of energy required to raise the temperature of unit mass of the substance by $1\,^\circ {\text{C}}$.

The change I temperature required to heat the copper is given as.

$\Delta T = {T_f} - {T_i}$  

Substitute the final and initial values of temperature.

$\begin{aligned}\Delta T &= 60 - 23\,^\circ {\text{C}} \\&= {\text{37}}\,^\circ {\text{C}} \\\end{aligned}$  

The expression for the amount of heat required to raise the temperature of the copper is.

 $\boxed{Q = mC\Delta T}$

Here, $Q$ is the amount of heat required, $m$ is the mass of the substance, $C$ is the specific heat capacity and $\Delta T$ is the change in temperature required.

Substitute the values in above expression.

$\begin{aligned}Q&=368\,\text{g}\times(0.0920\,\text{cal/g}\cdot{^\circ\text{C}})\times(37\,{^\circ}\text{C})\\&=1252.6\,\text{cal}\end{aligned}$  

Thus, the amount of heat required to raise the temperature of the copper from $23\,{\text{^\circ C}}$ to $60\,{\text{^\circ C}}$ is $\boxed{1252.6\,{\text{cal}}}$.

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Answer Details:

Grade: High School

Chapter: Heat transfer

Subject: Physics

Keywords:  Heat energy, raise the temperature, substance, copper, unit mass, heat required, specific heat, mass, 23 C to 60 C, capacity.

Answer 2

The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.

Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal