Question

Five grams of 20°C water is poured onto a 6-g block of ice that has an initial temperature of -15°C.
What is the final temperature if the effect of the surroundings can be neglected? .( specific heat of
ice = 0.50 cal/g.0C)

Answer 1

To solve this problem, we can use the heat gained/lost formula:
Q = mcΔT
Where:
Q = heat gained/lost (in calories)
m = mass (in grams)
c = specific heat capacity (in cal/g°C)
ΔT = change in temperature (in °C)
First, we need to calculate the amount of heat required to heat the ice from -15°C to its melting point (0°C). This can be calculated as:
Q_ice = m * c_ice * ΔT_ice
Q_ice = 6g * 0.50 cal/g°C * (0 - (-15))°C
Q_ice = 6g * 0.50 cal/g°C * 15°C
Q_ice = 45 cal
Next, we need to calculate the amount of heat required to melt the ice into water at 0°C. This can be calculated as:
Q_melt = m * heat of fusion
Q_melt = 6g * 80 cal/g
Q_melt = 480 cal
The total heat required to raise the temperature of the ice block from -15°C to water at 0°C is:
Q_total = Q_ice + Q_melt
Q_total = 45 cal + 480 cal
Q_total = 525 cal
Now, we need to calculate the temperature change of the 5g of 20°C water as it cools down to the final temperature. This can be calculated as:
Q_water = m * c_water * ΔT_water
Q_water = 5g * 1 cal/g°C * (20 - TF)°C
Q_water = 5g * 1 cal/g°C * (20 - TF)°C
Q_water = 5 cal * (20 - TF) cal
Q_water = 100 cal - 5TF cal
Since the heat lost by the water must equal the heat gained by the ice:
Q_water = Q_total
100 - 5TF = 525
Therefore, solving for TF:
5TF = 100 - 525
5TF = -425
TF = -425/5
TF = -85°C
The final temperature is -85°C.