A 8.52 kg block of ice has a temperature of -17.9o C. The pressure is one atmosphere. The block absorbs 3.780×106 J of heat. What is the final temperature of the liquid water? (Give your answer in oC, but enter only the numerical portion--oC is implied)
I need the equation and answer Thanks!
To solve this problem, we need to use the specific heat capacity formula and the equation for the heat absorbed by the substance:
Q = m * C * ΔT
Where:
Q = Heat absorbed by the substance (in joules)
m = Mass of the substance (in kilograms)
C = Specific heat capacity of the substance (in joules per kilogram per degree Celsius)
ΔT = Change in temperature (in degree Celsius)
In this case, we know the mass of the ice (m = 8.52 kg) and the heat absorbed (Q = 3.780×10^6 J). We also need to know the specific heat capacity of ice and water to calculate the change in temperature. Here are their values:
Specific heat capacity of ice (C_ice) = 2.093 J/g·°C
Specific heat capacity of water (C_water) = 4.186 J/g·°C
To convert the mass of the ice block from grams to kilograms, we use the conversion factor: 1 kg = 1000 g.
Let's calculate the change in temperature (ΔT) using the formula:
Q = m * C * ΔT
Rearranging the formula to solve for ΔT:
ΔT = Q / (m * C)
First, for the ice:
ΔT_ice = Q / (m_ice * C_ice)
Converting mass and specific heat capacity to grams and J/g·°C:
m_ice = 8.52 kg * 1000 g/kg = 8520 g
C_ice = 2.093 J/g·°C
Substituting these values into the equation:
ΔT_ice = 3.780×10^6 J / (8520 g * 2.093 J/g·°C)
Simplifying:
ΔT_ice = 3.780×10^6 J / (1.783 × 10^4 J/°C)
ΔT_ice ≈ 212.13 °C
Since the ice started at -17.9 °C, the final temperature (T_final) of the liquid water is given by:
T_final = -17.9 °C + ΔT_ice
T_final ≈ -17.9 °C + 212.13 °C
T_final ≈ 194.23 °C
Therefore, the final temperature of the liquid water is approximately 194.23 °C.