Question

A metal smith pours 2.8 kg of lead shot at 94.5^oC into 1.1 kg of water at 38.9^oC in an isolated container. What is the final temperature of the mixture? Give your answer to one decimal place in units of K.

C_water = 4.19 kJ/(kg*K), C_lead = 0.129 kJ/(kg*K)

2.If heat is added to a material, the temperature must increase.

true or false

3.

C_water = 4.19 kJ/(kg*K)

C_ice = 2.06 kJ/(kg*K),

L_fusion=334 kJ/kg

L_vaporization=2260 kJ/kg

How much heat (in kJ) is required to convert 0.8 kg of ice at -40^oC to ice at 0^oC? Give your answer to one decimal place.

4.

C_water = 4.19 kJ/(kg*K)

C_ice = 2.06 kJ/(kg*K),

L_fusion=334 kJ/kg

L_vaporization=2260 kJ/kg

How much heat (in kJ) is required to convert 0.7 kg of ice at 0^oC to water at 99^oC? Give your answer as a whole number.  

Answer 1

1. Use the principle of method of mixtures Heat lost by hot bodies Heat gain by cold bodies. Here lead shot is the hot body and water is the cold body Let the equilibrium temperature be at t. Then heat lost by lead shot omea (d-) Heat gain by water ea mC Then (2.8)(0.129)(94.5-)-(1.1) (4.19)(t-38.9) (11)(419) 38.9) (2.8)(00.129)

94.5- 12.76(t-38.9) 94.5-t 12.76t-496.364 94.5+496.364 -12.76t+t 590.864 13.76t 590.864 13.76 t42.9°C Hence, the final temperature of the mixture is t-42.9°C 2. If heat is added to the material, temperature need not be increased. In case of change of state, when heat is added, material changes its state without increasing its temperature. 3. Heat required to convert ice at -40° C to ice at 0°C is 0-mr.С.,(0-(-40)) (0.8)(206) (40) 65.9 k

4. Heat required to convert ice at 0°C to water at 99°C is Om nmer (99-0) -(0.7) (334)+(0.7) (4.19) (99) - 233.8+290.367 524 kJ