Question

How much heat, in joules and in calories, must be removed from 1.75 mol of water to lower its temperature from 25 degree celsius to 15 degree celsius?

Answer 1

How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from -23.0 °C to -1.0 °C?

Answer 2

Thus heat that must be removed from 1.75 mol of water to lower its temperature from 25 degree celsius to 15 degree celsius is 1318 Joules and 315 calories

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=-m\times c\times \Delta T$

Q = Heat removed= ?

m= mass of substance = $moles\times {\text{molar Mass}}=1.75mol\times 18g/mol=31.5g$

c = specific heat capacity = $4.184 J/g^0C$

Initial temperature of the water = $T_i$ = 25.0°C

Final temperature of the water = $T_f$  = 15.0°C

Change in temperature ,$\Delta T=T_f-T_i=(15-25)^0C=10^0C$

Putting in the values, we get:

$q=-31.5g\times 4.184J/g^0C\times -10^0C=$

$q=1318J$

$q=1318\times 0.24=315cal$    (4.184J=1cal)

Thus heat that must be removed from 1.75 mol of water to lower its temperature from 25 degree celsius to 15 degree celsius is 1318 Joules and 315 calories

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