First slide

Calorie

Question

A piece of lead is dropped from a height of 210 m. If 50 % of its striking energy is converted into heat calculate the rise in temperature (Specific heat of lead = 0.03 KCal / kg, J = 4200J/KCal and g = 10 m/s²)

Moderate

A

30°C
B

21°C
C

${(\frac{25}{3})}^{0} c$
D

${(\frac{55}{3})}^{0} C$

Solution

$\begin{array}{l} (m \times 10 \times 210) \times \frac{50}{100} = m \times 0 .03 \times 4200 \times Δθ \\ \Rightarrow Δθ = {\frac{25}{3}}^{\circ} C \end{array}$

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