Question

How much heat energy is needed to change 55 g
of ice at 0°C
into water at 25°C
? Use 4.2Jg°C
for the specific heat of water. Use 334 J/g
for the heat of fusion of water.(1 point)
Responses

2.4×104 J
2.4×104 J
- no response given

1.8×104 J
1.8×104 J
- no response given

1.3×104 J
1.3×104 J
- no response given

5.8×103 J
5.8×103 J
- incorrect

Answer 1

To find out how much heat energy is needed to change 55 g of ice at 0°C to water at 25°C, we need to consider two steps:

1. **Melting the ice at 0°C to water at 0°C.**
2. **Heating the water from 0°C to 25°C.**

**Step 1: Melting the ice**

We can calculate the energy required to melt the ice using the heat of fusion formula:

\[
q_1 = m \cdot L_f
\]

where:
- \(m\) = mass of the ice = 55 g
- \(L_f\) = heat of fusion of water = 334 J/g

Calculating \(q_1\):

\[
q_1 = 55 \, \text{g} \cdot 334 \, \text{J/g} = 18370 \, \text{J}
\]

**Step 2: Heating the water**

Now we need to calculate the energy required to heat the water from 0°C to 25°C:

\[
q_2 = m \cdot c \cdot \Delta T
\]

where:
- \(c\) = specific heat of water = 4.2 J/g°C
- \(\Delta T\) = change in temperature = 25°C - 0°C = 25°C

Calculating \(q_2\):

\[
q_2 = 55 \, \text{g} \cdot 4.2 \, \text{J/g°C} \cdot 25 \, \text{°C} = 5775 \, \text{J}
\]

**Total heat energy required:**

Now, adding \(q_1\) and \(q_2\):

\[
q_{total} = q_1 + q_2 = 18370 \, \text{J} + 5775 \, \text{J} = 24145 \, \text{J}
\]

Rounding it gives us:

\[
q_{total} \approx 2.4 \times 10^4 \, \text{J}
\]

Thus, the heat energy needed to change 55 g of ice at 0°C into water at 25°C is approximately **2.4×10² J**.