Home
/
Physics
/
1. What is the temperature change of a 130 g plece of aluminum whose specific heat is 2.89J/gcdot ^circ C after 1000 J of heat energy is applied? 2. How much energy is required to heat 22.5g of water from 34^circ C to 45^circ C 3. What is the specific heat of water if a beaker containing water weighs 150.0 g and is heated with a 16000 J of heat energy?The temperature of water changed from 22.0^circ C to 23.0^circ C 4. What is the mass of the beaker containing water if the temperature of water is 100^circ C and the specific heat of water is 4.18J/gOC after 1800J of heat energy is applied?

Question

1. What is the temperature change of a 130 g plece of aluminum whose specific heat is 2.89J/gcdot ^circ C after 1000 J of heat energy is applied? 2. How much energy is required to heat 22.5g of water from 34^circ C to 45^circ C 3. What is the specific heat of water if a beaker containing water weighs 150.0 g and is heated with a 16000 J of heat energy?The temperature of water changed from 22.0^circ C to 23.0^circ C 4. What is the mass of the beaker containing water if the temperature of water is 100^circ C and the specific heat of water is 4.18J/gOC after 1800J of heat energy is applied?

1. What is the temperature change of a 130 g plece of aluminum whose specific heat is 2.89J/gcdot ^circ C after 1000 J of heat energy is applied? 2. How much energy is required to heat 22.5g of water from 34^circ C to 45^circ C 3. What is the specific heat of water if a beaker containing water weighs 150.0 g and is heated with a 16000 J of heat energy?The temperature of water changed from 22.0^circ C to 23.0^circ C 4. What is the mass of the beaker containing water if the temperature of water is 100^circ C and the specific heat of water is 4.18J/gOC after 1800J of heat energy is applied?

expert verifiedVerification of experts

Answer

4.3247 Voting
avatar
JenaProfessional · Tutor for 6 years

Answer

### 1. \( 2.66 \: ^\circ C \)<br />### 2. \( 1034.85 \: J \)<br />### 3. \( 10.67 \: J/g \cdot^\circ C \)<br />### 4. \( 430.14 \: g \)

Explain

## Step1: Identify the formula for temperature change <br />The formula for calculating temperature change is \( q = mc\Delta T \), where:<br />- \( q \) is the heat energy added (in Joules),<br />- \( m \) is the mass of the substance (in grams),<br />- \( c \) is the specific heat capacity (in \( \mathrm{J/g \cdot ^\circ C} \)),<br />- \( \Delta T \) is the change in temperature (in \( \mathrm{^\circ C} \)).<br /><br />We need to reorganize this to solve for \( \Delta T \): <br />\[ \Delta T = \frac{q}{mc} \]<br /><br />## Step2: Calculate temperature change for aluminum <br />Given:<br />- Mass \( m = 130 \) g<br />- Specific heat capacity \( c = 2.89 \) J/g·°C<br />- Heat energy \( q = 1000 \) J<br /><br />Using the formula:<br />\[ \Delta T = \frac{1000}{130 \cdot 2.89} = \frac{1000}{375.7} \approx 2.66 \: ^\circ C \]<br /><br />## Step3: Determine energy required to heat water <br />Given:<br />- Mass \( m = 22.5 \) g<br />- Initial temperature \( T_i = 34^\circ C \)<br />- Final temperature \( T_f = 45^\circ C \)<br />- Specific heat capacity \( c = 4.18 \) J/g·°C (standard value for water)<br /><br />Calculate temperature change first:<br />\[ \Delta T = T_f - T_i = 45 - 34 = 11 \: ^\circ C \]<br /><br />Using the formula \( q = mc\Delta T \):<br />\[ q = 22.5 \cdot 4.18 \cdot 11 \approx 1034.85 \: J \]<br /><br />## Step4: Calculate the specific heat of water <br />Given:<br />- Mass of water \( m = 150.0 \) g<br />- Heat energy \( q = 1600.0 \) J<br />- Temperature change \( \Delta T = 23.0 - 22.0 = 1 \: ^\circ C \)<br /><br />Using the rearranged specific heat formula \( c = \frac{q}{m\Delta T} \):<br />\[ c = \frac{1600.0}{150.0 \cdot 1} = 10.67 \: J/g·^\circ C \]<br /><br />## Step5: Calculate the mass of the beaker containing water <br />Given:<br />- Temperature change \( \Delta T = 100^\circ C - T_o = 1 \: ^\circ C \)<br />- Specific heat \( c = 4.18 \: J/g \cdot^\circ C \)<br />- Heat energy \( q = 1800 \: J \)<br /><br />Using the formula rearranged for mass \( m \):<br />\[ m = \frac{q}{c \Delta T} = \frac{1800}{4.18 \cdot 1} \approx 430.14 \: g \]
Click to rate:

Hot Questions

More x