Question

You pour out 276 g of tea, (specific heat c = 1.00 c a l g...

You pour out 276 g of tea, (specific heat c = 1.00 c a l g K, same as H2O), pretty hot to start, T i =349 K. However, you also want to sweeten it with some honey, mass m h =152g, right out of the refrigerator at temperature T i = 277 K. Honey's specific heat is c = 0.64 c a l g K.

Calculate the thermal equilibrium of the tea with honey.

Type in the equilibrium temperature to the nearest Kelvin. E.g., if your answer it 233.588 K, then type in 234.

Homework Answers

Answer #1

According to principle of mixtures,

Heat lost by tea= Heat gained by Honey

Mt- Mass of tea

Mh- mass of honey

Ct- specific heat capacity of tea

Ch- specific heat capacity of honey

Tt- temperature of tea

Th - temperature of honey

T- equilibrium temperature

Mt* Ct*(Tt- T) = Mh*Ch*(T-Th)

Mt*Ct*Tt- Mt*Ct*T= Mh*Ch*T- Mh*Ch*Th

T= (Mt*Ct*Tt + Mh*Ch*Th)/(Mt*Ct+ Mh*Ch)

T= ((276*1*349)+ (152*0.64*277))/((276*1)+(152*0.64)) = 123270.56/373.28= 330.23

So Equilibrium temperature is 330 K

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