Home
/
Physics
/
The student places 0.5 kg of potato into a pan of water. During cooking the temperature of the potato increases from 20^circ C to 100^circ C The specific heat capacity of the potato is 3400J/kg^circ C Calculate the change in thermal energy of the potato. Use the equation: change nge in thermal energy=masstimes spadfic heat capacitytimes temparature dhe __

Question

The student places 0.5 kg of potato into a pan of water.
During cooking the temperature of the potato increases from
20^circ C to 100^circ C
The specific heat capacity of the potato is
3400J/kg^circ C
Calculate the change in thermal energy of the potato.
Use the equation:
change
nge in thermal energy=masstimes spadfic heat capacitytimes temparature dhe
__

The student places 0.5 kg of potato into a pan of water. During cooking the temperature of the potato increases from 20^circ C to 100^circ C The specific heat capacity of the potato is 3400J/kg^circ C Calculate the change in thermal energy of the potato. Use the equation: change nge in thermal energy=masstimes spadfic heat capacitytimes temparature dhe __

expert verifiedVerification of experts

Answer

4.2261 Voting
avatar
EliVeteran · Tutor for 9 years

Answer

After performing the multiplication, we find that the change in thermal energy of the potato is \( \Delta Q = 136000 \mathrm{~J} \).

Explain

## Step 1: <br />Identify the given values from the problem. We are given the mass of the potato \( m = 0.5 \mathrm{~kg} \), the specific heat capacity of the potato \( c = 3400 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C} \), and the change in temperature \( \Delta T = 100^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 80^{\circ} \mathrm{C} \).<br /><br />## Step 2: <br />Use the given equation to calculate the change in thermal energy. The equation is:<br /><br />### \( \Delta Q = m \times c \times \Delta T \)<br /><br />## Step 3: <br />Substitute the given values into the equation:<br /><br />### \( \Delta Q = 0.5 \mathrm{~kg} \times 3400 \mathrm{~J} / \mathrm{kg}{ }^{\circ} \mathrm{C} \times 80^{\circ} \mathrm{C} \)
Click to rate:

Hot Questions

More x