A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water so that the sphere is at 100 °C. It is then immediately transferred to a 0.14 kg copper calorimeter containing 0.25 kg water at 20 °C. The temperature of water rises and attains a steady-state at 23 °C. The specific heat capacity of aluminium is:

(Given that: Specific heat capacity of copper calorimeter = $0.386\times {10}^3 \mathrm{J} {\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$ and 

the specific heat capacity of water $\left({\mathrm{s}}_{\mathrm{w}}\right) = 4.18\times {10}^3 \mathrm{J} {\mathrm{kg}}^{-1} {\mathrm{K}}^{-1}$)

1.   $1.811 \mathrm{kJ} {\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$

2.   $1.911 \mathrm{kJ} {\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$

3.   $0.811 \mathrm{kJ} {\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$

4.   $0.911 \mathrm{kJ} {\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$

When \(0.15\) kg of ice at \(0^\circ \text{C}\) is mixed with \(0.30\) kg of water at \(50^\circ \text{C}\) in a container, the resulting temperature is \(6.7^\circ \text{C}.\)
The heat of fusion of ice is: (\(S_{\text{water}}=4186\) J kg^-1 K^-1)
1. \( 3.43 \times 10^4\) Jkg^-1
2. \( 3.34 \times 10^4\) Jkg^-1
3. \( 3.34 \times 10^5\) Jkg^-1
4. \(4.34 \times 10^5\) Jkg^-1

The heat required to convert 3 kg of ice at –12 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure is:

(Given, the specific heat capacity of ice = $2100 \mathrm{J} {\mathrm{kg}}^{-1}{\mathrm{K}}^{-1}$, the specific heat capacity of water = $4186 {\mathrm{Jkg}}^{-1}{\mathrm{K}}^{-1}$, the latent heat of fusion of ice = $3.35\times {10}^5 \mathrm{J} {\mathrm{kg}}^{-1}$
and the latent heat of steam = $2.256\times {10}^6 \mathrm{J} {\mathrm{kg}}^{-1}$.)

1.  $9.1\times {10}^7 \mathrm{J}$

2.  $8.1\times {10}^6 \mathrm{J}$

3.  $9.1\times {10}^6 \mathrm{J}$

4.  $8.1\times {10}^7 \mathrm{J}$