QuestionJune 17, 2025

A 32.1 g sample of an unknown substance is heated to 89.0^circ C and then was plunged into 55.0 mL of 20.0^circ C water in a coffee cup calorimeter. The temperature of the water rose to 28.1^circ C Calculate the specific heat of the metal. Specific heat of water=4.184J/g^circ C Density of water=1.00g/mL. a) 0.528J/g^circ C b) 0.606J/g^circ C C) 0.713J/g^circ C d) 0.953J/g^circ C e) 0.216J/g^circ C

A 32.1 g sample of an unknown substance is heated to 89.0^circ C and then was plunged into 55.0 mL of 20.0^circ C water in a coffee cup calorimeter. The temperature of the water rose to 28.1^circ C Calculate the specific heat of the metal. Specific heat of water=4.184J/g^circ C Density of water=1.00g/mL. a) 0.528J/g^circ C b) 0.606J/g^circ C C) 0.713J/g^circ C d) 0.953J/g^circ C e) 0.216J/g^circ C
A 32.1 g sample of an unknown substance is heated to 89.0^circ C and then was plunged into 55.0 mL of 20.0^circ C water in a coffee cup calorimeter. The temperature of the water rose to 28.1^circ C Calculate the specific heat of the metal. Specific heat of water=4.184J/g^circ C Density of water=1.00g/mL. a) 0.528J/g^circ C b) 0.606J/g^circ C C) 0.713J/g^circ C d) 0.953J/g^circ C e) 0.216J/g^circ C

Solution
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Answer

0.953 \, \text{J/g}^{\circ}\text{C} Explanation 1. Calculate the mass of water Mass of water = Volume × Density = 55.0 \, \text{mL} \times 1.00 \, \text{g/mL} = 55.0 \, \text{g} 2. Calculate heat absorbed by water Use **q = m \cdot c \cdot \Delta T** for water: q_{\text{water}} = 55.0 \, \text{g} \times 4.184 \, \text{J/g}^{\circ}\text{C} \times (28.1^{\circ}\text{C} - 20.0^{\circ}\text{C}) = 55.0 \times 4.184 \times 8.1 3. Calculate heat lost by metal Heat lost by metal = Heat gained by water = q_{\text{water}} = 1860.732 \, \text{J} 4. Calculate specific heat of the metal Use **q = m \cdot c \cdot \Delta T** for metal: 1860.732 \, \text{J} = 32.1 \, \text{g} \times c \times (89.0^{\circ}\text{C} - 28.1^{\circ}\text{C}) Solve for c: c = \frac{1860.732}{32.1 \times 60.9}

Explanation

1. Calculate the mass of water<br /> Mass of water = Volume × Density = $55.0 \, \text{mL} \times 1.00 \, \text{g/mL} = 55.0 \, \text{g}$<br /><br />2. Calculate heat absorbed by water<br /> Use **$q = m \cdot c \cdot \Delta T$** for water: <br /> $q_{\text{water}} = 55.0 \, \text{g} \times 4.184 \, \text{J/g}^{\circ}\text{C} \times (28.1^{\circ}\text{C} - 20.0^{\circ}\text{C}) = 55.0 \times 4.184 \times 8.1$<br /><br />3. Calculate heat lost by metal<br /> Heat lost by metal = Heat gained by water = $q_{\text{water}} = 1860.732 \, \text{J}$<br /><br />4. Calculate specific heat of the metal<br /> Use **$q = m \cdot c \cdot \Delta T$** for metal:<br /> $1860.732 \, \text{J} = 32.1 \, \text{g} \times c \times (89.0^{\circ}\text{C} - 28.1^{\circ}\text{C})$<br /> Solve for $c$: $c = \frac{1860.732}{32.1 \times 60.9}$
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