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How much heat is released when 275 g of water cools from 85.2°C to 38.4 °C?

Answer 1

Ellie Andrews

#Q = -53,796.6 J#

The equation for heat is
#Q = m(t_f -t_i)C_p#

Q = heat in Joules
m = mass in grams
#t_f# = final temp
#t_i# = initial temp
#C_p# = Specific Heat Capacity

For this problem
#Q = ???#
#m = 275 g#
#t_f = 38.4^oC#
#t_i = 85.2^oC#
#C_p = 4.18 j/(g^oC)#

#Q = 275g(38.4^oC -85.2^oC)4.18J/(g^oC)#
#Q = 275cancel(g)( -46.8^ocancelC))4.18J/(cancel(g^oC))#

#Q = -53,796.6 J#

The negative Joule value means heat is being lost in an exothermic reaction.

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Answer 2

Cooper Adams

To calculate the heat released, you can use the formula:

Q = mcΔT

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

Given: m = 275 g (mass of water) ΔT = (85.2°C - 38.4°C) = 46.8°C (change in temperature)

The specific heat capacity of water is approximately 4.18 J/g°C.

Substitute the values into the formula:

Q = (275 g) * (4.18 J/g°C) * (46.8°C)

Calculate the value of Q.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.