A lab assistant pours 330 g of water at $45 ^ { \circ } \mathrm { C }$ into an 855-g aluminum container that is at an initial temperature of $10 ^ { \circ } \mathrm { C }$ The specific heat of aluminum is $900 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and that of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

Question

A lab assistant pours  330 g of water at $45 ^ { \circ } \mathrm { C }$ into an  855-g aluminum container that is at an initial temperature of $10 ^ { \circ } \mathrm { C }$ The specific heat of aluminum is $900 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ and that of water is $4186 \mathrm {~J} / \mathrm { kg } \cdot \mathrm { K }$ What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

Quizplus · Accepted Answer

Explanation:The heat lost by the water is equal to the heat gained by the aluminum container.$$Q_{water} = Q_{aluminum}$$$$m_{water} c_{water} (T_{initial,water} - T_{final}) = m_{aluminum} c_{aluminum} (T_{final} - T_{initial,aluminum})$$$$(0.330 	ext{ kg})(4186 	ext{ J/kg} \cdot 	ext{K})(45 ^ { \circ } \mathrm { C } - T_{final}) = (0.855 	ext{ kg})(900 	ext{ J/kg} \cdot 	ext{K})(T_{final} - 10 ^ { \circ } \mathrm { C })$$$$141054 - 1393.8T_{final} = 771.75T_{final} - 7717.5$$$$21822.5 = 2164.55T_{final}$$$$T_{final} = 10.08 ^ { \circ } \mathrm { C } \approx 10 ^ { \circ } \mathrm { C }$$

A Lab Assistant Pours 330 G of Water At 45