calories need to melt 60.0 of ice at 0 and to warm the liquid to 65 (two steps)

Same problem but different numbers.

http://www.jiskha.com/display.cgi?id=1315786222

To find out the total calories needed to melt 60.0g of ice at 0°C and then warm the resulting liquid to 65°C, we need to calculate the calories required for each step separately and then add them together.

Step 1: Melting the ice
To melt the ice, we need to provide enough heat to change its state from solid ice at 0°C to liquid water at 0°C. The amount of heat required to melt a substance is given by the formula:

q = m * ΔH_fus

where
q is the heat in calories,
m is the mass of the substance in grams, and
ΔH_fus is the heat of fusion (or the enthalpy of fusion) of the substance.

For water, ΔH_fus is generally accepted to be 79.7 calories/gram. So, to determine the heat required to melt the ice, we calculate:

q1 = m1 * ΔH_fus
= 60.0g * 79.7 cal/g
≈ 4782 cal

Therefore, it would take around 4782 calories to melt 60.0g of ice at 0°C.

Step 2: Warming the liquid
After the ice is melted, we need to warm the resulting liquid water from 0°C to 65°C. The amount of heat required to raise the temperature of a substance is given by the formula:

q = m * C * ΔT

where
q is the heat in calories,
m is the mass of the substance in grams,
C is the specific heat capacity of the substance, and
ΔT is the change in temperature.

For water, the specific heat capacity is approximately 1 calorie/gram°C. So, to calculate the heat required to warm the liquid water, we use:

q2 = m2 * C * ΔT
= 60.0g * 1 cal/g°C * (65°C - 0°C)
= 60.0 * 1 * 65 cal
= 3900 cal

Therefore, it would take around 3900 calories to warm the liquid water from 0°C to 65°C.

Total Calories Required:
To find the total calories required for both steps, we add the heat required for each step:

Total calories = q1 + q2
= 4782 cal + 3900 cal
≈ 8682 cal

Hence, it would take approximately 8682 calories to melt 60.0g of ice at 0°C and then warm the resulting liquid water to 65°C.