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How much energy is needed to heat 36.0 grams of ice, H2O, from

-10.5°C to water at 22.5°C?
State Specific Heat
Solid 2.108 J/g °C
Liquid 4.182 J/g °C
Gas 1.996 J/g °C

User Shabeer
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2 Answers

4 votes

Final answer:

To heat 36.0 grams of ice from -10.5°C to water at 22.5°C, 4113.324 J of energy is needed. The calculation involves two steps: heating the ice from -10.5°C to its melting point and then heating the water from its melting point to 22.5°C. The specific heats of ice and liquid water are used in the calculations.

Step-by-step explanation:

To calculate the amount of energy needed to heat the ice from -10.5°C to water at 22.5°C, we need to consider two steps: heating the ice from -10.5°C to its melting point, and then heating the water from its melting point to 22.5°C.

For the first step, we use the specific heat of ice, which is 2.108 J/g °C. The temperature change is 0°C - (-10.5°C) = 10.5°C, and the mass of ice is 36.0 g. The energy required for this step is calculated using the formula Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat, and ΔT is the temperature change. Therefore, Q1 = (36.0 g)(2.108 J/g °C)(10.5°C) = 790.584 J.

For the second step, we use the specific heat of liquid water, which is 4.182 J/g °C. The temperature change is 22.5°C - 0°C = 22.5°C, and the mass of water is 36.0 g. The energy required for this step is calculated using the same formula, Q = mcΔT. Therefore, Q2 = (36.0 g)(4.182 J/g °C)(22.5°C) = 3322.74 J.

The total energy needed to heat the ice from -10.5°C to water at 22.5°C is the sum of Q1 and Q2, which is Q1 + Q2 = 790.584 J + 3322.74 J = 4113.324 J.

User KimHafr
by
8.4k points
3 votes

Final answer:

To heat 36.0 grams of ice from -10.5°C to water at 22.5°C, a total of 123145.13 J of energy is needed.

Step-by-step explanation:

To determine the amount of energy needed to heat the ice, we will break down the process into two steps:

First, we need to calculate the energy required to raise the temperature of the ice from -10.5°C to 0°C.

Second, we need to calculate the energy required to melt the ice at 0°C and raise the temperature of the resulting water from 0°C to 22.5°C.

For the first step, we use the specific heat of solid ice, which is 2.108 J/g °C:

Energy = mass × specific heat × temperature change

Energy = 36.0 g × 2.108 J/g °C × (0°C - (-10.5°C))

Energy = 36.0 g × 2.108 J/g °C × 10.5°C

Energy = 794.884 J

For the second step, we need to calculate the energy required to melt the ice and raise the temperature of the resulting water:

Energy = energy to melt the ice + energy to raise the temperature of the water

Energy to melt the ice = mass × heat of fusion

Energy to melt the ice = 36.0 g × 79.9 cal/g × (4.184 J/cal)

Energy to melt the ice = 119910.816 J

Energy to raise the temperature of the water = mass × specific heat of liquid water × temperature change

Energy to raise the temperature of the water = 36.0 g × 4.182 J/g °C × (22.5°C - 0°C)

Energy to raise the temperature of the water = 3439.43 J

Adding the two energies together, we get:

Total energy = 794.884 J + 119910.816 J + 3439.43 J

Total energy = 123145.13 J

User BlakeTNC
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8.7k points