A100 g sample of h2o liquid at 22°c absorbs 8360 jewels of heat what will be the final temperature of the water

42°C

Explanation:

The formula relating the heat (q) absorbed or released by an object, its mass (m), and the temperature change (ΔT) is

q = mCΔT     Divide both sides by mC and switch

ΔT = q/(mC)

where

C = the specific heat capacity of the object

1. Calculate the change in temperature

q = 8360 J

m = 100 g

C = 4.184 J·°C⁻¹g⁻¹

ΔT = 8360/(100 × 4.184)

ΔT = 8360/418.4

ΔT = 19.98 °C

2. Calculate the final temperature

ΔT = T₂ -T₁     Add T₁ to each side and switch

T₂ = T₁ + ΔT

where T₂ and T₁ are the final and initial temperatures

T₁ = 22 °C

ΔT = 19.98 °C

T₂ = 22 + 19.98

T₂ = 42 °C

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