Answer to Question #168026 in General Chemistry for azyyy

Question

Please with steps

Compare the number of calories absorbed when 668 g of ice at 0°C is changed to liquid water at 38°C with the number of calories absorbed when 668 g of liquid water is warmed from 0°C to 38°C.

amount of heat=......................cal

Answer 1

For water

Q = mc $\Delta T$

$\Delta T$ = 38⁰C - 0⁰C = 38⁰C

Specific heat of water = 4.186 joule/gram ⁰C

Q = 668 × 4.186 × 38 = 106,256.424 joules

106,256.425÷4.184 = 25,395.89 calories

For ice

Q = mc $\Delta T$

$\Delta T$ = 38⁰C - 0⁰C = 38⁰C

Specific heat of ice = 2.01 J/g ⁰C

Q = 668×2.01×38 = 51,021.84 joules

51,021.84÷4.184 = 12,194.51 calories