Question

A block of a certain material is heated to a temperature of $500^\circ\text{C}$ and then placed on a large ice block. If $455\text{ kg}$ of ice melts, find the mass of the block. Specific heat of the material is $39\text{ J/g}^\circ\text{C}$ and heat of fusion of water is $335\text{ J/g}$.

• A. $1.455\text{ kg}$
• B. $2.5\text{ kg}$
• C. $0.67\text{ kg}$
• D. $2.67\text{ kg}$

Hint:

Always ensure your units for specific heat and latent heat match up. Here, both were given in terms of per gram ($/\text{g}$), so converting the mass of ice into grams first makes the algebraic steps straightforward and less prone to decimal placement errors.

Answer 1

The Correct Option is B

Concept: According to the principle of calorimetry, in an isolated system, the total heat lost by hotter bodies is equal to the total heat gained by cooler bodies: \[ Q_{\text{lost}} = Q_{\text{gained}} \] The heat released by the block as it cools down from $T$ to $0^\circ\text{C}$ is given by $Q = m \cdot s \cdot \Delta T$. The heat taken up by the ice block to change its state from solid to liquid at a fixed temperature of $0^\circ\text{C}$ is given by $Q = m_{\text{ice}} \cdot L_f$.

Step 1: List values and equate expressions under consistent units.
Let's choose grams ($\text{g}$) and Joules ($\text{J}$) for our quantities:
• Temperature change of the block, $\Delta T = 500^\circ\text{C} - 0^\circ\text{C} = 500^\circ\text{C}$
• Specific heat capacity of the material, $s = 0.39\text{ J/g}^\circ\text{C}$
• Mass of ice melted, $m_{\text{ice}} = 1.455\text{ kg} = 1455\text{ g}$
• Latent heat of fusion, $L_f = 335\text{ J/g}$ Let the mass of the block be $m$ (in grams). \[ Q_{\text{lost}} = m \times s \times \Delta T \] \[ Q_{\text{gained}} = m_{\text{ice}} \times L_f \] Applying calorimetry: \[ m \times 0.39 \times 500 = 1455 \times 335 \]

Step 2: Solve for the mass of the block ($m$).
\[ m \times 195 = 487425 \] \[ m = \frac{487425}{195} = 2500\text{ g} \] Converting back to kilograms: \[ m = \frac{2500}{1000} = 2.5\text{ kg} \]