QuestionJuly 8, 2025

A 55.3 g sample of brass is put into a calorimeter (see sketch at right) that contains 100.0 g of water. The brass sample starts off at 97.1^circ C and the temperature of the water starts off at 17.0^circ C When the temperature of the water stops changing it's 21.4^circ C The pressure remains constant at 1 atm. Calculate the specific heat capacity of brass according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

A 55.3 g sample of brass is put into a calorimeter (see sketch at right) that contains 100.0 g of water. The brass sample starts off at 97.1^circ C and the temperature of the water starts off at 17.0^circ C When the temperature of the water stops changing it's 21.4^circ C The pressure remains constant at 1 atm. Calculate the specific heat capacity of brass according to this experiment. Be sure your answer is rounded to the correct number of significant digits.
A 55.3 g sample of brass is put into a calorimeter (see sketch at right) that contains 100.0 g of water. The brass sample starts off at 97.1^circ C and the temperature of the water starts off at 17.0^circ C When the temperature of the water stops changing it's 21.4^circ C The pressure remains constant at 1 atm. Calculate the specific heat capacity of brass according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

Solution
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Answer

0.375 \, \frac{J}{g\cdot ^{\circ }C} Explanation 1. Calculate heat gained by water Use q = m \cdot c \cdot \Delta T. For water, m = 100.0 \, \text{g}, c = 4.18 \, \frac{J}{g\cdot ^{\circ }C}, \Delta T = 21.4 - 17.0 = 4.4^{\circ}C. So, q_{\text{water}} = 100.0 \cdot 4.18 \cdot 4.4. 2. Calculate heat lost by brass Heat lost by brass is equal to heat gained by water: q_{\text{brass}} = q_{\text{water}}. 3. Calculate specific heat capacity of brass Use q = m \cdot c \cdot \Delta T. For brass, m = 55.3 \, \text{g}, \Delta T = 97.1 - 21.4 = 75.7^{\circ}C. Solve for c: c = \frac{q_{\text{brass}}}{m \cdot \Delta T}.

Explanation

1. Calculate heat gained by water<br /> Use $q = m \cdot c \cdot \Delta T$. For water, $m = 100.0 \, \text{g}$, $c = 4.18 \, \frac{J}{g\cdot ^{\circ }C}$, $\Delta T = 21.4 - 17.0 = 4.4^{\circ}C$. So, $q_{\text{water}} = 100.0 \cdot 4.18 \cdot 4.4$.<br />2. Calculate heat lost by brass<br /> Heat lost by brass is equal to heat gained by water: $q_{\text{brass}} = q_{\text{water}}$.<br />3. Calculate specific heat capacity of brass<br /> Use $q = m \cdot c \cdot \Delta T$. For brass, $m = 55.3 \, \text{g}$, $\Delta T = 97.1 - 21.4 = 75.7^{\circ}C$. Solve for $c$: $c = \frac{q_{\text{brass}}}{m \cdot \Delta T}$.
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