Question

Certain amount of heat is given to 100 g of copper to increase its temperature by 21 $^{\circ}$C. if the same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat capacity of copper = 400 Jkg$^{-1} K^{-1}$ and that for water =4200Jkg$^{-1}ZK^{-1}$

• A. 4$^{\circ}$C
• B. 5.25$^{\circ}$C
• C. 8$^{\circ}$C
• D. 6$^{\circ}$C

Answer 1

The Correct Option is A

The amount of heat supplied is given by the
relation $ \, \, \, \, \, Q=mc \Delta T$
Here , m = 100g = 0. 1 kg,c = 400J/ kg -K
$\, \, \, \, \, \, \, \, \Delta T =21 K \, for \, copper$
$Thus , \, \, \, \, \, \, \, \, Q = 0.1 \times 400 \times 21 = 840J$
Hence , $ \, \, \, \, \, 840=0.05 \times 4200 \times \Delta T$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \Delta T =4^{\circ}C$

Answer 2

To solve this problem, we'll use the concept of specific heat capacity. The specific heat capacity (\(c\)) is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or \(1^{\circ} \text{C}\)).

The formula to calculate the heat (\(Q\)) added to a substance is:

\[ Q = mc\Delta T \]

where:
- \(m\) is the mass of the substance,
- \(c\) is the specific heat capacity,
- \(\Delta T\) is the change in temperature.

Step 1: Calculate the heat required to raise the temperature of copper

Given:
- Mass of copper, \(m_{\text{Cu}} = 100 \, \text{g} = 0.1 \, \text{kg}\)
- Specific heat capacity of copper, \(c_{\text{Cu}} = 400 \, \text{J/kg} \cdot \text{K}\)
- Temperature change for copper, \(\Delta T_{\text{Cu}} = 21^{\circ} \text{C}\)

\[ Q = m_{\text{Cu}} \cdot c_{\text{Cu}} \cdot \Delta T_{\text{Cu}} \]

\[ Q = 0.1 \, \text{kg} \cdot 400 \, \text{J/kg} \cdot \text{K} \cdot 21 \, \text{K} \]

\[ Q = 0.1 \cdot 400 \cdot 21 \]

\[ Q = 840 \, \text{J} \]

So, the amount of heat required to raise the temperature of the copper is 840 J.

 Step 2: Calculate the rise in temperature for water using the same amount of heat

Given:
- Mass of water, \(m_{\text{H2O}} = 50 \, \text{g} = 0.05 \, \text{kg}\)
- Specific heat capacity of water, \(c_{\text{H2O}} = 4200 \, \text{J/kg} \cdot \text{K}\)
- Heat added, \(Q = 840 \, \text{J}\)

We need to find the temperature change \(\Delta T_{\text{H2O}}\):

\[ Q = m_{\text{H2O}} \cdot c_{\text{H2O}} \cdot \Delta T_{\text{H2O}} \]

Rearranging to solve for \(\Delta T_{\text{H2O}}\):

\[ \Delta T_{\text{H2O}} = \frac{Q}{m_{\text{H2O}} \cdot c_{\text{H2O}}} \]

\[ \Delta T_{\text{H2O}} = \frac{840 \, \text{J}}{0.05 \, \text{kg} \cdot 4200 \, \text{J/kg} \cdot \text{K}} \]

\[ \Delta T_{\text{H2O}} = \frac{840}{0.05 \cdot 4200} \]

\[ \Delta T_{\text{H2O}} = \frac{840}{210} \]

\[ \Delta T_{\text{H2O}} = 4 \, \text{K} \]

Therefore, the rise in temperature for water when 840 J of heat is added is \(4^{\circ} \text{C}\).
So correct Answer is option A \(4^{\circ} \text{C}\).