How many calories does it take to heat 10 grams of ice to 29 degrees C?

Question

 please help

Answer 1

 

m = mass of ice = 10 g

 

You need to look up the latent heat of fusion of water, L_f.  I found a table with

 

L_f = 333.7 kJ/kg = 333.7 J/g

 

You also need the specific heat of water, c.  I found a table with 

 

c = 4.23 kJ/(kg°C) = 4.23 J/(g°C)

 

The ice melts at a constant temperature of 0°C.  The heat required to melt it is 

 

Q₁ = mL_f

 

After it is fully melted, additional heat will raise its temperature.

 

T_i = initial temperature = 0°C

T_f = final temperature = 29°C

 

The heat required to raise the temperature of the melted ice from T_i to T_f is

 

Q₂ = mc(T_f - T_i)

 

The total heat required is

 

Q = Q₁ + Q₂

 

Q = mL_f + mc(T_f - T_i) = m[L_f + c(T_f - T_i)]

 

Q = (10 g) [(333.7 J/g) + (4.23 J/(g°C)) (29°C - 0°C)] ≅ 4564 J 

 

You want the answer in calories,

 

1 cal = 4.18 J

 

Q = (4564 J) / (4.18 J/cal) ≅ 1092 cal