Question

how much energy would be lost by 23 g of water if it was heated until it was 78 degrees celsius and then allow to cool down to 27 degrees celsius

Answer 1

Q = -4903.14 j

Step-by-step explanation:

Given data:

Mass of water = 23 g

Initial temperature = 78°C

Final temperature = 27°C

Heat lost = ?

Solution:

Specific heat capacity of water is 4.18 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 27°C - 78°C

ΔT = -51°C

Q = 23 g × 4.18 J/g.°C × -51°C

Q = -4903.14 j (negative sign shows heat is released)