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The temperature of a sample of water increases from 20°C to 46°C as it absorbs 5650 calories of heat. What is the mass of the sample? (Specific heat of the water is 1.0 Cal/g °C

Sagot :

luana
[tex]Formula\ for\ specific\ heat:\\\\ c=\frac{\Delta\ Q}{m*\Delta t}\\ c-\ specific\ heat\\ m-\ mass\\ \Delta Q-\ change\ of\ calories\\ \Delta T-\ change\ of\ temperature\\\\ 1=\frac{5650}{m(46-20)}\\\\ 1=\frac{5650}{26m}\ \ \ |*26m\\\\ 26m=5650\ \ \ |:26\\\\ m=217,3\ grams\\\\ Sample\ weight\ 217,3\ grams.[/tex]
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