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

[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]