We drop a cube of ice into a glass of water. The mass of the cube of ice is 33.1 g and its initial temperature is −10.2∘C. The mass of the water is 251 g and its initial temperature is 19.7∘C. What is the final temperature of the water after all of the ice has melted?

The final temperature (T) will be;

16.23°c

Explanation: To find the final temperature of any mixture of substance with an initial temperature use the formula;

M1c(T-T1) + M2c(T-T2)=0

c is the specific heat capacity of the two different substance, but because ice and water are the same, we assume c to be 1

T= final temperature of the mixture

T1= initial temperature of the ice= -10.2°c

T2= initial temperature of the water= 19.7°c

M1= mass of ice= 33.1g

M2= mass of water= 251g

Using the formula above

33.1(T-(-10.2)) + 251(T-19.7)=0

Solving out the bracket

33.1T + 337.62 + 251T - 4944.7 = 0

Collecting like terms to both side of the equation and solving

33.1T + 251T = 4944.7 - 33.62

284.1T = 4607.08

T = 4607.08÷284.1 = 16.23°c

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.