A student carries out an investigation where she mixes 200 g of liquid H at a te

Question

A student carries out an investigation where she mixes 200 g of liquid H at a temperature of 80 degrees Celcius with 100 g of liquid K at a temperature of 20 degrees Celcius ending up with a 300 g mixutre at a temperature of 40 degrees Celcius. Six variations to this intial situation are depicted below.

A: 200 g of liquid H at a temperature of 90 degrees Celcius

100 g of liquid K at a temperature of 40 degrees Celcius

B: 100 g of liquid H at a temperature of 40 degrees Celcius

200 g of liquid K at a temperature of 90 degrees Celcius

C: 200 g of liquid H at a temperature of 90 degrees Celcius

200 g of liquid K at a temperature of 60 degrees Celcius

D: 100 g of liquid H at a temperature of 90 degrees Celcius

200 g of liquid K at a temperature of 30 degrees Celcius

E: 400 g of liquid H at a temperature of 30 degrees Celcius

100 g of liquid K at a temperature of 90 degrees Celcius

F: 200 g of liquid H at a temperature of 20 degrees Celcius

200 g of liquid K at a temperature of 90 degrees Celcius

Rank cases A-F based on the final temperatures of the mixture.

***You do not need to calculate each value, just show the general way to approach this problem as I cannot seem to figure it out without having the energies in joules.

Explanation / Answer

General way of approach is that , we can calculate the Heat Released as

Q = M1c1deltaT+ M2C2DeltaT

We Know the specific Heats for H and K and also the Temparatures and Change in Temparature and masses , Hence Q can be calculated

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