What mass of water will change its temperature by 30.0 °C when 800.0 Joules of heat is added to it? The specific heat of water is 4.18 J/g°c
m=800.0J4.18J/g°C×30°Cm=\frac {800.0J} {4.18J/g°C×30°C}m=4.18J/g°C×30°C800.0J
m=6.38g
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