Studdy AI Logo
Math  /  Data & Statistics

QuestionA 150.5−g150.5-\mathrm{g} sample of a metal at 75.4∘C75.4^{\circ} \mathrm{C} is added to 150.5 gHH2O150.5 \mathrm{~g} \mathrm{H} \mathrm{H}_{2} \mathrm{O} at 15.3∘C15.3^{\circ} \mathrm{C}. The temperature of the water rises to 18.8∘C18.8^{\circ} \mathrm{C}. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat capacity of water is 4.18 J/∘C⋅g4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}.
Specific heat capacity == □\square J/∘C⋅g\mathrm{J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}

Studdy Solution

STEP 1

1. The mass of the metal is 150.5 g 150.5 \, \text{g} .
2. The initial temperature of the metal is 75.4∘C 75.4^\circ \text{C} .
3. The mass of the water is 150.5 g 150.5 \, \text{g} .
4. The initial temperature of the water is 15.3∘C 15.3^\circ \text{C} .
5. The final temperature of both the metal and the water is 18.8∘C 18.8^\circ \text{C} .
6. The specific heat capacity of water is 4.18 J/g∘C 4.18 \, \text{J/g}^\circ \text{C} .
7. Heat lost by the metal equals heat gained by the water.

STEP 2

1. Calculate the heat gained by the water.
2. Calculate the heat lost by the metal.
3. Solve for the specific heat capacity of the metal.

STEP 3

Calculate the heat gained by the water using the formula:
qwater=mwater⋅cwater⋅ΔTwaterq_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}
where: - mwater=150.5 g m_{\text{water}} = 150.5 \, \text{g} - cwater=4.18 J/g∘C c_{\text{water}} = 4.18 \, \text{J/g}^\circ \text{C} - ΔTwater=18.8∘C−15.3∘C=3.5∘C \Delta T_{\text{water}} = 18.8^\circ \text{C} - 15.3^\circ \text{C} = 3.5^\circ \text{C}
qwater=150.5⋅4.18⋅3.5q_{\text{water}} = 150.5 \cdot 4.18 \cdot 3.5
qwater=2203.565 Jq_{\text{water}} = 2203.565 \, \text{J}

STEP 4

Since the heat lost by the metal equals the heat gained by the water:
qmetal=−qwater=−2203.565 Jq_{\text{metal}} = -q_{\text{water}} = -2203.565 \, \text{J}

STEP 5

Calculate the specific heat capacity of the metal using the formula:
qmetal=mmetal⋅cmetal⋅ΔTmetalq_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}}
where: - mmetal=150.5 g m_{\text{metal}} = 150.5 \, \text{g} - ΔTmetal=18.8∘C−75.4∘C=−56.6∘C \Delta T_{\text{metal}} = 18.8^\circ \text{C} - 75.4^\circ \text{C} = -56.6^\circ \text{C}
Rearrange to solve for cmetal c_{\text{metal}} :
cmetal=qmetalmmetal⋅ΔTmetalc_{\text{metal}} = \frac{q_{\text{metal}}}{m_{\text{metal}} \cdot \Delta T_{\text{metal}}}
cmetal=−2203.565150.5⋅−56.6c_{\text{metal}} = \frac{-2203.565}{150.5 \cdot -56.6}
cmetal=2203.5658518.3c_{\text{metal}} = \frac{2203.565}{8518.3}
cmetal≈0.259 J/g∘Cc_{\text{metal}} \approx 0.259 \, \text{J/g}^\circ \text{C}
The specific heat capacity of the metal is:
0.259 J/g∘C \boxed{0.259 \, \text{J/g}^\circ \text{C}}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord