Dissolving some potassium bromide in 200cm3 of water leads to a decrease in temperature of 3oC. What will the heat released be?

Question

Serenity Allen · Accepted Answer

The answer is #q = -2510J#.

The amount of heat released can be calculated using the equation

#q = m_(H_2O) * c_(H_2O) * DeltaT#, where

#m_(H_2O)# - the mass of water;
#c_(H_2O)# -water's specific heat ( #4.18 J/(g^@C)#);
#DeltaT# - the change in temperature measured as #T_(fi nal) - T_(i nitial)#.

A decrease in temperature will determine a negative #DeltaT#, since this is equivalent to a lower final temperature (I'm assuming that the decrease in temperature was #3^@C#, not #30^@C#).

We can use water's density of approximately #1.00g/(cm^3)# to determine its mass

#rho = m/V -> m_(H_2O) = rho * V = 1.00 g/(cm^3) * 200.0 cm^3 = 200.0g#

Therefore,

#q = m * c * DeltaT = 200.0g * 4.18 J/(g^@C) * (-3.00^@C) = -2510J#

The water lost energy in the form of heat equal to #2510J#.

Ellie Andrews · Answer

undefined The heat released can be calculated using the equation:

Q = mcΔT

Where:
Q = heat released (in joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)

Given that the change in temperature is -3°C and the mass of water is 200g (since 1 cm^3 of water = 1 g), we can plug in these values to calculate the heat released:

Q = (200g) * (4.18 J/g°C) * (-3°C)
Q ≈ -2508 J

Therefore, the heat released is approximately -2508 joules.