If 52 j of heat is added to an aluminum can with a mass of 18.3 g , what is its temperature change? specific heat capacity of aluminum is 0.903 j/ g ∘ c.

I believe the temperature change will be; 3.147 °C
Heat supplied by system is equivalent to the heat gained by aluminium metal.The heat supplied in this case is 52 Joules, thus
Heat energy gained by aluminium = mcΔT, where m is the mass of aluminium in Kg, c is the specific heat capacity of aluminium and ΔT is the change in temperature.
Therefore, 52 = 18.3 × 0.903 × ΔT
 52 = 16.5249 ΔT 
ΔT = 3.147 
Therefore, the temperature change is 3.147 degrees celsius

3.147 °C


Explanation:

1) Data:


Q = 52 J

m = 18.3 g

Cs = 0.903 j / g ∘c.

ΔT = ?


2) Formula


Q = m x Cs x ΔT


3) Calculations:

i) Solve for ΔT


ΔT = Q / [m x Cs}


ii) Plug in the data


ΔT = 52 J / [ 18.3 g x 0.903 J/g°C] = 3.147°C

Mass of water is smething that you froma drug test or sum girl ya

Ithink d hope that