If you place a 37.8g ice cube (0 degrees C) into a 200gram container of water at 21.0 degrees C, what is the coolest temperature the water will reach? Please show your work. Thanks! :)
-
Latent heat of ice = 80 calorie/g
Specific heat of water = 1cal/g
37.8g at 0°C requires 37.8*80= 3024calories s to form water at 0°C
200g water will give out 200*21 = 4200calories if it cools from 21 to 0°C
Excess calories (4200 -3024) =1176 calories are used to raise the temperature of
(200+ 37.8) = 237.8 g of water
Raise in temperature = 1176/ 237.8 = 4.95°C
The maximum temperature is 4.95 °C
=====================================
OR
37.8g at 0°C requires 37.8*80= 3024calories s to form water at 0°C
37.8g of water requires 37.8*t calories to raise its temperature from 0°C to t°C
Total calories needed =3024 + 37.8*t
----------------------------
200 g of water gives out 200*(21-t) ° calories when it cools from 21 to t°C
---------------------------
Equating the two
3024 + 37.8*t = 4200 - 200t
237.8 t =1176
t = 4.95°C
The maximum temperature is 4.95 °C
=====================================
Specific heat of water = 1cal/g
37.8g at 0°C requires 37.8*80= 3024calories s to form water at 0°C
200g water will give out 200*21 = 4200calories if it cools from 21 to 0°C
Excess calories (4200 -3024) =1176 calories are used to raise the temperature of
(200+ 37.8) = 237.8 g of water
Raise in temperature = 1176/ 237.8 = 4.95°C
The maximum temperature is 4.95 °C
=====================================
OR
37.8g at 0°C requires 37.8*80= 3024calories s to form water at 0°C
37.8g of water requires 37.8*t calories to raise its temperature from 0°C to t°C
Total calories needed =3024 + 37.8*t
----------------------------
200 g of water gives out 200*(21-t) ° calories when it cools from 21 to t°C
---------------------------
Equating the two
3024 + 37.8*t = 4200 - 200t
237.8 t =1176
t = 4.95°C
The maximum temperature is 4.95 °C
=====================================