561.2 J of heat was added to 8.00 g of water. The final temperature of the water was measured to be 64.2°C. What was the initial temperature of the water? (specific heat capacity of water = 4.184 J/g°C)
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q = m * SH * ΔT
ΔT = q / (m * SH)
64.2 - x = 561.2 / (8.00 * 4.184)
64.2 - x = 16.8
x = 47.4
ΔT = q / (m * SH)
64.2 - x = 561.2 / (8.00 * 4.184)
64.2 - x = 16.8
x = 47.4