Question

26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated cup.What will the new water temperature be?

Answer 1

Answer:

The new water temperature is 26.4 °C

Explanation:

Given;

mass of copper, $M_{cu}$ = 26 g = 0.026 kg

temperature of copper, t = 300 °C

volume of water, V = 120 mL = 0.12 L

temperature of water, t = 21 °C

density of water, ρ = 1 kg/L

mass of water = density x volume

mass of water = (1 kg/L) x 0.12 L = 0.12 kg

heat lost by copper = heat gained by water

Both copper and water reach final temperature, T

Heat gained by water, $Q_w$ = $m_w$cΔθ = $m_w C(T - t)$

$Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)$

Heat lost by copper is given by;

$Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)$

$Q_{cu} = Q_w$

504(T- 21) = 10.01(300 - T)

504 T - 10584 = 3003 - 10.01 T

504 T + 10.01 T= 3003 + 10584

514.01 T = 13587

T = (13587) / 514.01

T = 26.4 °C

Therefore, the new water temperature is 26.4 °C