You have 750 g of water at 10°C in a large insulated beaker. How much boiling water at 100°C must you add to this beaker so that the final
temperature of the mixture will be 75°C? For water, c = 4.19 x 103 J/(kg • K).
naman ang kanilang a movementvou

Correct  m₂ = 1,950 g

Explanation:

When mixing the liquid, in this case water, at different temperatures, the warmer liquid will cool and the cooler will warm up until their temperatures equalize.

The amount of heat that the second fluid (water) releases is equal to the amount of heat that the first fluid (water) receives in this spontaneous process.

Q₁ = Q₂

Q₁ = m₁ c (t - t₁)    and Q₂ = m₂ c (t₂ - t)  

where are they:  m₁ = 750 g, t₁ = 10°C, t₂ = 100°C, t = 75°C, m₂ = ?

m₁ c (t - t₁)  = m₂ c (t₂ - t) \ : c

Since water is the liquid in question, it has the same specific heat capacity. Therefore, we will divide both sides of the equation by c and get:

m₁ (t - t₁)  = m₂ (t₂ - t) =>  m₂ = m₁ (t - t₁) / (t₂ - t) =>

m₂ = 750 · (75 - 10) / (100 - 75) => m₂ = 750 · 65 / 25 = 1,950 g

m₂ = 1,950 g

God is with you!!!

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.