Question

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.

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