Question

In a real calorimeter, most of the heat released is absorbed by water but certain amount of heat is also absorbed by metal and surrounding water tank.

Calorimeter, absorbs 5 cal/ºC. If 50 gm water of 60°C is mixed with calorimeter's original 50 gm water at 25ºC. What will be the final temperature?

Answer 1

$\frac{125}{3} \, \text{°C}$

Answer 2

The principle of calorimetry states that heat lost by hot water equals heat gained by cold water and the calorimeter.

Heat lost by hot water: $Q_{lost} = m_h c_w (T_h - T_f)$ Heat gained by cold water: $Q_{gained\_water} = m_c c_w (T_f - T_c)$ Heat gained by calorimeter: $Q_{gained\_cal} = C_{cal} (T_f - T_c)$

Given: $m_h = 50 \, \text{gm}$, $T_h = 60 \, \text{°C}$ $m_c = 50 \, \text{gm}$, $T_c = 25 \, \text{°C}$ $C_{cal} = 5 \, \text{cal/°C}$ Assume $c_w = 1 \, \text{cal/gm°C}$.

Equation: $50 \times 1 \times (60 - T_f) = 50 \times 1 \times (T_f - 25) + 5 \times (T_f - 25)$ $50(60 - T_f) = (50 + 5)(T_f - 25)$ $3000 - 50T_f = 55(T_f - 25)$ $3000 - 50T_f = 55T_f - 1375$ $4375 = 105T_f$ $T_f = \frac{4375}{105} = \frac{125}{3} \, \text{°C}$