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1 year ago

Hello, I have a physics exam tomorrow and I don’t understand how to do these questions. I know the basic concepts. Could anyone do a worked solution for me as once I see how to do it I will understand it. That would help a lot. Thanks!!

1) 200g of metal is heated in a flame to a temperature of 600 degrees Cand dropped into boiling liquid. It is found that 20g of the liquid vaporises. If the specific heat capacity of the metal is 0.5J/g degrees C and the boiling point of the liquid is 100 degrees C , find the specific heat capacity of the liquid.

2) a piece of ice, mass 100g, is at -4 degrees C and is supplied with 40kJ of energy. Assuming all energy is absorbed by the ice what is its final temperature? (Shc of ice = 2100 J/kg degrees C) (shc of water 4200) (slh(f) ice = 336000 j/kg)

3) dry steam from a boiler at 100 degrees C is passed into a copper can of mass 200g of which contains a mixture of 450g of water and 50g of ice all at 0 degrees C. The steam supply is cut off when the temperature in the can reaches 20 degrees C after thorough stirring. Subsequent weighing reveals that 23g of steam has been condensed. Assuming there is no heat loss find a value for the specific Latent heat of vaporisation of steam. (Shc water 4.2 j/gdegrees C) (Shc of copper 0.4 j/g degreees C) ( slh of ice = 336 j/g)

1) 200g of metal is heated in a flame to a temperature of 600 degrees Cand dropped into boiling liquid. It is found that 20g of the liquid vaporises. If the specific heat capacity of the metal is 0.5J/g degrees C and the boiling point of the liquid is 100 degrees C , find the specific heat capacity of the liquid.

2) a piece of ice, mass 100g, is at -4 degrees C and is supplied with 40kJ of energy. Assuming all energy is absorbed by the ice what is its final temperature? (Shc of ice = 2100 J/kg degrees C) (shc of water 4200) (slh(f) ice = 336000 j/kg)

3) dry steam from a boiler at 100 degrees C is passed into a copper can of mass 200g of which contains a mixture of 450g of water and 50g of ice all at 0 degrees C. The steam supply is cut off when the temperature in the can reaches 20 degrees C after thorough stirring. Subsequent weighing reveals that 23g of steam has been condensed. Assuming there is no heat loss find a value for the specific Latent heat of vaporisation of steam. (Shc water 4.2 j/gdegrees C) (Shc of copper 0.4 j/g degreees C) ( slh of ice = 336 j/g)

Hi, I'm a second year physics and philosophy student at the University of Nottingham.

The wording for 1 confuses me slightly but I can certainly get you going with 2 and 3

2) Here you want to set up an equation with the energy supplied to the ice on one side (the 40kJ), and the energy gained by the ice on the other. I'll use C_w or C_i to denote whether I am talking about the SHC of water or ice respectively. Similarly for change in temperature (ΔT_i for the ice's change in temperature and ΔT_w for water). Because of the info given, we can assume that the ice will go up in temperature, change state, and heat up a bit as water before we even calculate anything.

Using this, on the RHS of the equation, we would have the energy for the 100g of ice to get up to boiling point become water, E = m(C_i)(ΔT_i), the energy for the ice to become water, E = mL, and then for that same mass of (now) water to heat up to the final temperature of the water, E = m(C_w)(ΔT_w)

This becomes: 40,000 = m(C_i)(ΔT_i) + mL + m(C_w)(ΔT_w)

Because we know m, C_i, ΔT_i, L, and C_w, we can rearrange our equation to find ΔT_w as it is the only unknown. Then just sub in the values and get ΔT_w.

3) Here, as there is no energy loss, the energy lost from the steam condensing to water and dropping in temperature is equal to the energy gained from the copper, water and ice reaching a temperature of 20 degrees C. Here C_c, C_w are the SHC of copper and water with similar notation for the mass, change in temperature, and latent heat (where necessary) of the copper, ice, water, and steam.

Again setting up an equation for energy loss is important. Once we do this, we can rearrange and solve for L, the SLH of steam.

On the LHS we would have the energy loss from the 100g of steam in condensing to water and then dropping from 100 degrees C to 20 degrees C. E = (M_s)(L_c) + (M_s)(C_w)(ΔT_s)

On the RHS we have the energy gained by the copper and 450g of water getting to 20 degrees C, E = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w), as well as the energy gained by ice in changing state and then getting to 20 degrees C, E = (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

This becomes: (M_s)(L_c) + (M_s)(C_w)(ΔT_s) = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w) + (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

We know all the mass values, temperature change values, and SHCs so we can sub them in. We also know the SLH of ice leaving the only thing we don't know as the SLH of the steam. To to get this just rearrange and solve. You could check your answer is close by checking online after working it out. Be careful with unit conversions though!

Hope this helps

The wording for 1 confuses me slightly but I can certainly get you going with 2 and 3

2) Here you want to set up an equation with the energy supplied to the ice on one side (the 40kJ), and the energy gained by the ice on the other. I'll use C_w or C_i to denote whether I am talking about the SHC of water or ice respectively. Similarly for change in temperature (ΔT_i for the ice's change in temperature and ΔT_w for water). Because of the info given, we can assume that the ice will go up in temperature, change state, and heat up a bit as water before we even calculate anything.

Using this, on the RHS of the equation, we would have the energy for the 100g of ice to get up to boiling point become water, E = m(C_i)(ΔT_i), the energy for the ice to become water, E = mL, and then for that same mass of (now) water to heat up to the final temperature of the water, E = m(C_w)(ΔT_w)

This becomes: 40,000 = m(C_i)(ΔT_i) + mL + m(C_w)(ΔT_w)

Because we know m, C_i, ΔT_i, L, and C_w, we can rearrange our equation to find ΔT_w as it is the only unknown. Then just sub in the values and get ΔT_w.

3) Here, as there is no energy loss, the energy lost from the steam condensing to water and dropping in temperature is equal to the energy gained from the copper, water and ice reaching a temperature of 20 degrees C. Here C_c, C_w are the SHC of copper and water with similar notation for the mass, change in temperature, and latent heat (where necessary) of the copper, ice, water, and steam.

Again setting up an equation for energy loss is important. Once we do this, we can rearrange and solve for L, the SLH of steam.

On the LHS we would have the energy loss from the 100g of steam in condensing to water and then dropping from 100 degrees C to 20 degrees C. E = (M_s)(L_c) + (M_s)(C_w)(ΔT_s)

On the RHS we have the energy gained by the copper and 450g of water getting to 20 degrees C, E = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w), as well as the energy gained by ice in changing state and then getting to 20 degrees C, E = (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

This becomes: (M_s)(L_c) + (M_s)(C_w)(ΔT_s) = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w) + (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

We know all the mass values, temperature change values, and SHCs so we can sub them in. We also know the SLH of ice leaving the only thing we don't know as the SLH of the steam. To to get this just rearrange and solve. You could check your answer is close by checking online after working it out. Be careful with unit conversions though!

Hope this helps

Reply 2

1 year ago

Original post by JonahP553

Hi, I'm a second year physics and philosophy student at the University of Nottingham.

The wording for 1 confuses me slightly but I can certainly get you going with 2 and 3

2) Here you want to set up an equation with the energy supplied to the ice on one side (the 40kJ), and the energy gained by the ice on the other. I'll use C_w or C_i to denote whether I am talking about the SHC of water or ice respectively. Similarly for change in temperature (ΔT_i for the ice's change in temperature and ΔT_w for water). Because of the info given, we can assume that the ice will go up in temperature, change state, and heat up a bit as water before we even calculate anything.

Using this, on the RHS of the equation, we would have the energy for the 100g of ice to get up to boiling point become water, E = m(C_i)(ΔT_i), the energy for the ice to become water, E = mL, and then for that same mass of (now) water to heat up to the final temperature of the water, E = m(C_w)(ΔT_w)

This becomes: 40,000 = m(C_i)(ΔT_i) + mL + m(C_w)(ΔT_w)

Because we know m, C_i, ΔT_i, L, and C_w, we can rearrange our equation to find ΔT_w as it is the only unknown. Then just sub in the values and get ΔT_w.

3) Here, as there is no energy loss, the energy lost from the steam condensing to water and dropping in temperature is equal to the energy gained from the copper, water and ice reaching a temperature of 20 degrees C. Here C_c, C_w are the SHC of copper and water with similar notation for the mass, change in temperature, and latent heat (where necessary) of the copper, ice, water, and steam.

Again setting up an equation for energy loss is important. Once we do this, we can rearrange and solve for L, the SLH of steam.

On the LHS we would have the energy loss from the 100g of steam in condensing to water and then dropping from 100 degrees C to 20 degrees C. E = (M_s)(L_c) + (M_s)(C_w)(ΔT_s)

On the RHS we have the energy gained by the copper and 450g of water getting to 20 degrees C, E = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w), as well as the energy gained by ice in changing state and then getting to 20 degrees C, E = (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

This becomes: (M_s)(L_c) + (M_s)(C_w)(ΔT_s) = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w) + (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

We know all the mass values, temperature change values, and SHCs so we can sub them in. We also know the SLH of ice leaving the only thing we don't know as the SLH of the steam. To to get this just rearrange and solve. You could check your answer is close by checking online after working it out. Be careful with unit conversions though!

Hope this helps

The wording for 1 confuses me slightly but I can certainly get you going with 2 and 3

2) Here you want to set up an equation with the energy supplied to the ice on one side (the 40kJ), and the energy gained by the ice on the other. I'll use C_w or C_i to denote whether I am talking about the SHC of water or ice respectively. Similarly for change in temperature (ΔT_i for the ice's change in temperature and ΔT_w for water). Because of the info given, we can assume that the ice will go up in temperature, change state, and heat up a bit as water before we even calculate anything.

Using this, on the RHS of the equation, we would have the energy for the 100g of ice to get up to boiling point become water, E = m(C_i)(ΔT_i), the energy for the ice to become water, E = mL, and then for that same mass of (now) water to heat up to the final temperature of the water, E = m(C_w)(ΔT_w)

This becomes: 40,000 = m(C_i)(ΔT_i) + mL + m(C_w)(ΔT_w)

Because we know m, C_i, ΔT_i, L, and C_w, we can rearrange our equation to find ΔT_w as it is the only unknown. Then just sub in the values and get ΔT_w.

3) Here, as there is no energy loss, the energy lost from the steam condensing to water and dropping in temperature is equal to the energy gained from the copper, water and ice reaching a temperature of 20 degrees C. Here C_c, C_w are the SHC of copper and water with similar notation for the mass, change in temperature, and latent heat (where necessary) of the copper, ice, water, and steam.

Again setting up an equation for energy loss is important. Once we do this, we can rearrange and solve for L, the SLH of steam.

On the LHS we would have the energy loss from the 100g of steam in condensing to water and then dropping from 100 degrees C to 20 degrees C. E = (M_s)(L_c) + (M_s)(C_w)(ΔT_s)

On the RHS we have the energy gained by the copper and 450g of water getting to 20 degrees C, E = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w), as well as the energy gained by ice in changing state and then getting to 20 degrees C, E = (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

This becomes: (M_s)(L_c) + (M_s)(C_w)(ΔT_s) = (M_c)(C_c)(ΔT_c) + (M_w)(C_w)(ΔT_w) + (M_i)(L_c) + (M_i)(C_w)(ΔT_i)

We know all the mass values, temperature change values, and SHCs so we can sub them in. We also know the SLH of ice leaving the only thing we don't know as the SLH of the steam. To to get this just rearrange and solve. You could check your answer is close by checking online after working it out. Be careful with unit conversions though!

Hope this helps

Wow ! Thank you so much!! I think I’ve got the right answers now it makes so much more sense 😄

(edited 1 year ago)

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