A 200 g sample of hot water at \( 80^\circ \text{C} \) is poured into 300 g of cold water at \( 20^\circ \text{C} \) in an ideal calorimeter. What is the final temperature?

Solution:

Let final temperature be \( T \). Heat lost by hot water = Heat gained by cold water.

\( 200 \cdot 4.18 \cdot (80 - T) = 300 \cdot 4.18 \cdot (T - 20) \)

Simplifying:
\( 200(80 - T) = 300(T - 20) \)
\( 16000 - 200T = 300T - 6000 \)
\( 22000 = 500T \)
\( T = 44^\circ \text{C} \)

In a calorimeter of heat capacity 150 J/°C, 100 g of water at \( 90^\circ \text{C} \) is added to 200 g of water at \( 20^\circ \text{C} \). Find the final temperature assuming no heat loss.

Solution:

Let final temperature be \( T \). Heat lost = heat gained + heat absorbed by calorimeter

\( 100 \cdot 4.18 \cdot (90 - T) = 200 \cdot 4.18 \cdot (T - 20) + 150(T - 20) \)

Simplify:
\( 418(90 - T) = 836(T - 20) + 150(T - 20) \)
\( 37620 - 418T = 986(T - 20) \)
\( 37620 - 418T = 986T - 19720 \)
\( 57340 = 1404T \)
\( T \approx 40.8^\circ \text{C} \)

A 150 g piece of metal at \( 100^\circ \text{C} \) is placed into 250 g of water at \( 20^\circ \text{C} \). The final temperature is \( 24^\circ \text{C} \). Assuming no heat loss and water’s specific heat is 4.18 J/g°C, find the specific heat of the metal.

Solution:

Heat lost by metal = Heat gained by water

Let specific heat of metal be \( c \).
\( 150c(100 - 24) = 250 \cdot 4.18 \cdot (24 - 20) \)
\( 150c \cdot 76 = 250 \cdot 4.18 \cdot 4 \)
\( 11400c = 4180 \)
\( c = \frac{4180}{11400} \approx 0.367 \, \text{J/g°C} \)

A 50 g ice cube at \( 0^\circ \text{C} \) is added to 200 g of water at \( 30^\circ \text{C} \) in an ideal calorimeter. Find the final temperature. Assume latent heat of fusion for ice is \( 334 \, \text{J/g} \), and specific heat of water is \( 4.18 \, \text{J/g°C} \).

Solution:

Heat needed to melt ice: \( Q_{\text{melt}} = 50 \cdot 334 = 16700 \, \text{J} \)

Heat available from water cooling to \( 0^\circ \text{C} \):
\( Q_{\text{water}} = 200 \cdot 4.18 \cdot 30 = 25080 \, \text{J} \)

Since \( 25080 > 16700 \), all ice melts.
Remaining heat: \( 25080 - 16700 = 8380 \, \text{J} \)

Total water = 200 + 50 = 250 g
Final rise in temperature: \( \Delta T = \frac{8380}{250 \cdot 4.18} \approx 8.02^\circ \text{C} \)
Final temp ≈ \( 8.0^\circ \text{C} \)