THERMAL CYU 9

Solution 9

We apply the principle of calorimetry: Heat lost by water = Heat gained by the block.

Heat lost by water ($Q_w$): $$ Q_w = m_w s_w (T_i – T_f) $$ $$ Q_w = 0.5 \times 4200 \times (45 – T) $$

Heat gained by block ($Q_b$): Since specific heat $s$ depends on temperature ($s = s_0(1+k\theta)$), we must integrate. $$ Q_b = \int_{0}^{T} m_b s(\theta) d\theta = m_b \int_{0}^{T} s_0(1+k\theta) d\theta $$ $$ Q_b = m_b s_0 \left[ \theta + \frac{k\theta^2}{2} \right]_0^T = m_b s_0 \left( T + \frac{kT^2}{2} \right) $$ Given $m_b = 0.1$ kg, $s_0 = 4.2 \times 10^3$, $k = 0.1$.

Equating $Q_w = Q_b$: $$ 0.5 \times 4200 (45 – T) = 0.1 \times 4200 \left( T + \frac{0.1 T^2}{2} \right) $$ Cancel $4200$ from both sides: $$ 0.5 (45 – T) = 0.1 \left( T + 0.05 T^2 \right) $$ $$ 22.5 – 0.5T = 0.1T + 0.005T^2 $$ $$ T^2 + 120T – 4500 = 0 $$