Verified answer

A 2.5-kg water receives thermal energy of 36, 575 J. By how much its temperature increases? (Water has a specific heat capacity of 4,180 J/kg°C.)

Formula:

Q = mcΔT

=> (transpose to find ΔT)

=> ΔT = mc / Q

where:

m = 2.5 kg

c = 4,180 J/kg°C

Q = 36, 575 J

Calculation:

ΔT = [(2.5kg)(4180 J/kg°C)] / 36, 575 J

ΔT = (10,450 J/°C ) / 36, 575 J

ΔT = 0.2857142857 °C

ΔT = 0.2857 °C

Explanation :

• multply 2.5 kg with 4180J/kg°C

• then divide the 10,450 to 36,575J

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