Question

A 10kg iron bar(specific heat 011cal/gC°)at 80°C is placed on a block of ice.How much ice melts?

Answer 1

Answer on Question #72325, Physics / Mechanics | Relativity |

A 10 kg iron bar (specific heat 0.11 cal/g°C) at 80°C is placed on a block of ice. How much ice melts?

Solution

$$\begin{array}{l} m_{\text{ir}} = 10000 \text{ g} \\ C = 0.11 \text{ cal/g}{}^{\circ} \text{C} \\ \Delta t = (80{}^{\circ} \text{C} - 0{}^{\circ} \text{C}) = 80{}^{\circ} \text{C} \\ L = 81 \text{ cal/g} \\ m_{\text{ice}} - ? \\ Q_{\text{ir}} = Q_{\text{ice}}, \quad m_{\text{ir}} C \Delta t = m_{\text{ice}} L, \quad m_{\text{ice}} = (m_{\text{ir}} C \Delta t) / L. \end{array}$$

Finally

$$m_{\text{ice}} = (10000 \cdot 0.11 \cdot 80) / 81 = 1080 [\text{g}], \quad m_{\text{ice}} = 1080 \text{ g} = 1.08 \text{ kg}$$

Answer: 1.08 kg

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