Question

Ice to Steam via Water Due this Friday, Dec 4 at 11:59 pm (EST) How much heat is required to change a 49.4 g ice cube from ice at -13.7°C to water at 50°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C) 2.82×104 J You are correct. Your receipt no. is 150-2609 Help: Receipt Previous Tries How much heat is required to change a 49.4 g ice cube from ice at -13.7°C to steam at 120°C?

Answer 1

convert gram into Kg

49.4 g x (1 kg / 1000 g) = 0.0494 kg

caculate heat

1) The ice warms from -13.7ºC to its melting point, 0ºC.

Since this is a temperature change, use the formula q = mcΔT

q = (0.0494 kg)(2090 J/kgºC)(13.7ºC) = 1414.4702 J

2) The ice melts while the temperature holds steady at 0ºC.

Since this is a phase change, use the formula q = mL. Since this phase change is melting, use the heat of fusion for L, which is 3.34x10^5 J/kg.

q = (0.0494 kg)(3.34x10^5 J/kg) = 16499.6 J

3) The meltwater warms from 0ºC to 100ºC.

Another temperature change, but this time it's water (rather than ice), so you'll need to use the specific heat of water instead.

q = (0.0494 kg)(4186 J/kgºC)(100ºC) = 20678.84 J

4) The water boils while the temperature holds steady at 100ºC.

Another phase change. Use the heat of vaporization and the formula q = mL

q = (0.0494 kg)(2.26x10^6 J/kg) = 111644 J

5) The steam warms from 100ºC to 120ºC.

One last temperature change. The specific heat of steam is about 2010 J/kgºC (depending on the temperature and pressure).

q = (0.0494 kg)(2010 J/kgºC)(20ºC) = 1985.88 J

Adding the five heats together = 152222.79 J