A brief note on Specific Heat, Calorimetry and Change of State . Read it before a test. Try each question at the end before you open its answer.
The big ideas Specific heat : Q = m c Δ T Q = mc\,\Delta T Q = m c Δ T ; a body's heat capacity is m c mc m c . Water's c c c , 4186 J kg − 1 K − 1 4186\ \text{J kg}^{-1}\text{K}^{-1} 4186 J kg − 1 K − 1 , is large. Mixtures : heat lost = = = heat gained, if none escapes. Add the calorimeter's water equivalent. Latent heat : during melting or boiling the temperature stays steady and Q = m L Q = mL Q = m L . Pressure raises boiling points; evaporation cools. The facts at a glance Quantity Value c c c of water 4186 J kg − 1 K − 1 4186\ \text{J kg}^{-1}\text{K}^{-1} 4186 J kg − 1 K − 1 c c c of ice 2100 J kg − 1 K − 1 2100\ \text{J kg}^{-1}\text{K}^{-1} 2100 J kg − 1 K − 1 L f L_f L f of ice 3.34 × 10 5 J/kg 3.34 \times 10^{5}\ \text{J/kg} 3.34 × 1 0 5 J/kg L v L_v L v of water 2.26 × 10 6 J/kg 2.26 \times 10^{6}\ \text{J/kg} 2.26 × 1 0 6 J/kg Water equivalent m c c c c w \dfrac{m_cc_c}{c_w} c w m c c c
A worked example A cup of 0.25 kg 0.25\ \text{kg} 0.25 kg of tea at 85 ∘ C 85\ ^\circ\text{C} 85 ∘ C gets 0.050 kg 0.050\ \text{kg} 0.050 kg of milk at 5 ∘ C 5\ ^\circ\text{C} 5 ∘ C , then a 10 g 10\ \text{g} 10 g ice cube at 0 ∘ C 0\ ^\circ\text{C} 0 ∘ C . Treat everything as water.
Milk. T = ( 0.25 × 85 + 0.050 × 5 ) kg ∘ C 0.30 kg = 71.7 ∘ C T = \dfrac{(0.25 \times 85 + 0.050 \times 5)\ \text{kg}\,^\circ\text{C}}{0.30\ \text{kg}} = 71.7\ ^\circ\text{C} T = 0.30 kg ( 0.25 × 85 + 0.050 × 5 ) kg ∘ C = 71.7 ∘ C . Ice. 0.30 kg × 4186 J kg − 1 K − 1 × ( 71.7 ∘ C − T ) = 0.010 kg × 3.34 × 10 5 J/kg + 0.010 kg × 4186 J kg − 1 K − 1 T 0.30\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times (71.7\ ^\circ\text{C} - T) = 0.010\ \text{kg} \times 3.34 \times 10^{5}\ \text{J/kg} + 0.010\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1}\,T 0.30 kg × 4186 J kg − 1 K − 1 × ( 71.7 ∘ C − T ) = 0.010 kg × 3.34 × 1 0 5 J/kg + 0.010 kg × 4186 J kg − 1 K − 1 T . Answer. T = 66.8 ∘ C T = 66.8\ ^\circ\text{C} T = 66.8 ∘ C : the small cube cools the tea by about 5 K 5\ \text{K} 5 K , mostly through its latent heat. Practice with new numbers
A program to try The program works out the cup of tea.
Read it, then run it.
How many such ice cubes bring the tea down to 50 ∘ C 50\ ^\circ\text{C} 50 ∘ C ?
Mistakes to avoid The plain average for unequal masses. Weight each temperature. m c Δ T mc\,\Delta T m c Δ T during melting. Use m L mL m L : the temperature does not change. Forgetting that melted ice must warm up. Leaving out the calorimeter. It takes heat too. Not checking whether all the ice melts. If not, the answer is 0 ∘ C 0\ ^\circ\text{C} 0 ∘ C . Check yourself How much heat warms 0.80 kg 0.80\ \text{kg} 0.80 kg of water by 25 K 25\ \text{K} 25 K ? Stuck? Show a hint Show the answer0.5 kg 0.5\ \text{kg} 0.5 kg of water at 90 ∘ C 90\ ^\circ\text{C} 90 ∘ C and 1.5 kg 1.5\ \text{kg} 1.5 kg at 20 ∘ C 20\ ^\circ\text{C} 20 ∘ C are mixed. What is the final temperature? Stuck? Show a hint Show the answerHow much heat freezes 0.50 kg 0.50\ \text{kg} 0.50 kg of water at 0 ∘ C 0\ ^\circ\text{C} 0 ∘ C , and where does it go? Stuck? Show a hint Show the answerWhy does a pressure cooker cook faster? Stuck? Show a hint Show the answerWhy do we feel cold when we step out of a swimming pool on a windy day? Stuck? Show a hint Show the answer