Calculate the heat required to raise the temperature of 0.75 kg of water from 5c to 90.c?
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Given,
mass = 0.75
specific heat of water = 1
∆θ = T2 - T1 = 90 - 5 = 85°C
We will apply the formula :
Heat (q) = ms∆θ
q = 0.75 × 1 × 85
q = 63.75 J
Hope this helps..
✌
mass = 0.75
specific heat of water = 1
∆θ = T2 - T1 = 90 - 5 = 85°C
We will apply the formula :
Heat (q) = ms∆θ
q = 0.75 × 1 × 85
q = 63.75 J
Hope this helps..
✌
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