# Expert solution/answerSpecific heat of the water is 4.18 kJ/kg. °C.The expression for the heat transfer of the cold beverage is:Q = mc(T, -T)coldi cold10 m(4.18 kJ/kg - °C)(48°C –16°C)1 mL= (120 mL )(10 J= (16.0×10* kJ)1 kJ= 16 JI have used the expert formula above in calculating (d), and it happens to be section(c) answer is wrong but (d), is correct...may you please re-check for me where theproblem is with expert's solution (c.)Solution (d)Q = MhotC (Tf – T;i hot) 16 JI have used the expert formula above in calculating (d), and it happens to be section(c) answer is wrong but (d), is correct...may you please re-check for me where theproblem is with expert's solution (c.)Solution (d)Q = MhotC (Tf – Ti hot)[10-5m²-(120mL)(4. 18 kJ/kg.°C) (48-69°C)1ml(103('10.5 x 10*kJ) ()kJ= 10.5 JoulesThe formula above has worked for section (d)...I don't knowwhy sectionc is not correct.

Question
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Initial temperature hot water 69° C,
Final temperature after adding cold water 48° C.
Initial temperature of cold water is 16° C
Quantity warm water 175 ml

c)What is your calculated heat transfer (Q) of the cold beverage when 4 ounces (120 mL) was added to the hot beverage? Report your answer in Joules.

d) What is your calculated heat transfer (Q) of the hot beverage when 4 ounces (120 mL) of cold beverage was added? Report your answer in Joules.

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Step 1

(c)

Let cp be the specific heat capacity of water, mc be the mass of the cold water, mh be the mass of hot water, Tc be the temperature of cold water, Th be the temperature of hot water, Teq be the equilibrium temperature.

Write the expression for the heat transfer for the cold beverage.

Step 2

(d)

Write the expression for the heat ...

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