© Macmillan Learning To treat a burn on his hand, a person decides to place an ice cube on the burned skin. The mass of the ice cube is 12.7 g , and its initial temperature is $$ -11.9^{\circ} \mathrm{C} $$. The water resulting from the melted ice reaches the temperature of his skin, $$ 28.4^{\circ} \mathrm{C} $$. How much heat is absorbed by the ice cube and resulting water? Assume that all of the water remains in the hand. Constants for water can be found in this table. $$ q= $$ $$ \square $$

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Let 
$$
m = 12.7\,\text{g}, \quad T_i = -11.9\,^{\circ}\text{C}, \quad T_f = 28.4\,^{\circ}\text{C}.
$$
The process occurs in three steps:

1. **Warming the ice from $$-11.9\,^{\circ}\text{C}$$ to $$0\,^{\circ}\text{C}$$:**

The specific heat capacity of ice is 
   $$
   c_{\text{ice}} = 2.09 \, \frac{\text{J}}{\text{g}\cdot ^{\circ}\text{C}}.
   $$
   The temperature change is 
   $$
   \Delta T_1 = 0 - (-11.9) = 11.9\,^{\circ}\text{C}.
   $$
   Thus, the heat required is
   $$
   q_1 = m\, c_{\text{ice}}\, \Delta T_1 = 12.7\,\text{g} \times 2.09\, \frac{\text{J}}{\text{g}\cdot ^{\circ}\text{C}} \times 11.9\,^{\circ}\text{C}.
   $$
   Calculating, we get 
   $$
   q_1 \approx 12.7 \times 2.09 \times 11.9 \approx 315.8 \, \text{J}.
   $$

2. **Melting the ice at $$0\,^{\circ}\text{C}$$:**

The latent heat of fusion for water is 
   $$
   L_f = 334\, \frac{\text{J}}{\text{g}}.
   $$
   The heat required to melt the ice is
   $$
   q_2 = m\, L_f = 12.7\,\text{g} \times 334\, \frac{\text{J}}{\text{g}}.
   $$
   Calculating, we obtain 
   $$
   q_2 \approx 4241.8 \, \text{J}.
   $$

3. **Warming the resulting water from $$0\,^{\circ}\text{C}$$ to $$28.4\,^{\circ}\text{C}$$:**

The specific heat capacity of water is 
   $$
   c_{\text{water}} = 4.18 \, \frac{\text{J}}{\text{g}\cdot ^{\circ}\text{C}}.
   $$
   The temperature change is 
   $$
   \Delta T_3 = 28.4 - 0 = 28.4\,^{\circ}\text{C}.
   $$
   The heat required is
   $$
   q_3 = m\, c_{\text{water}}\, \Delta T_3 = 12.7\,\text{g} \times 4.18 \, \frac{\text{J}}{\text{g}\cdot ^{\circ}\text{C}} \times 28.4\,^{\circ}\text{C}.
   $$
   Calculating, we find 
   $$
   q_3 \approx 12.7 \times 4.18 \times 28.4 \approx 1508.0 \, \text{J}.
   $$

4. **Total Heat Absorbed:**

The net heat absorbed is the sum of the three steps:
   $$
   q = q_1 + q_2 + q_3 \approx 315.8 \, \text{J} + 4241.8 \, \text{J} + 1508.0 \, \text{J}.
   $$
   Therefore,
   $$
   q \approx 6065.6 \, \text{J}.
   $$

Thus, the heat absorbed by the ice cube and the resulting water is approximately 
$$
\boxed{6.07 \times 10^3 \, \text{J}}.
$$
The ice cube and resulting water absorb approximately 6,070 joules of heat.

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