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
4.3(344 votes)

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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