Question 962a7

Mar 10, 2014

It takes 3.17 times as many calories to change the ice to water at
37 °C as it does to warm the water to 37 °C.

a. Heat needed = heat to melt ice + heat to warm water

q_1 = q_2 + q_3 = mΔH_(fus) + mcΔT# =

254 g × 79.72 cal•g⁻¹ + 254 g × 1 cal.g⁻¹°C⁻¹ × 37 °C =

20 249 cal + 9398 cal = 29 647 cal

b. Heat to warm water = ${q}_{3}$ = 9398 cal

${q}_{1} / {q}_{3} = \frac{29 647 c a l}{9398 c a l}$ = 3.15

It requires 29 600 cal to change the ice to water and 9400 cal to warm the water. This is a ratio of 3.15:1.