How much heat (kJ) is needed to raise the temperature of 100.0 grams of water from 25.0°C to 50.0°C? Q=m x C x change in temp Q= ? m= 100.0g C=? tf 25.0 C - Ti 50.0 C trouble setting this up
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$\displaystyle Q = 4.8 \frac{J}{g C} * 100 \frac{g}{ml} * 25 C $ about 12 KJ
I came up with q= 100.0 x 4.18 x 25.0 10450. 10.5 kj
Originally Posted by Louisana1 I came up with q= 100.0 x 4.18 x 25.0 10450. 10.5 kj yea u are right, i should have put 4.18
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