How much heat is needed to raise the temperature of 100.0 grams of water from 25.0°C to 55.0°C? Specific heat capacity of water is 4.184 J/g°C.
Answer (1)
To calculate the amount of heat needed to raise the temperature of water, we can use the formula for heat transfer in specific heat capacity:
q = m ⋅ c ⋅ Δ T
Where:
q is the heat energy (in joules, J) we want to find.
m is the mass of the water (in grams, g).
c is the specific heat capacity (in J/g°C).
Δ T is the change in temperature (in °C).
Given:
m = 100.0 grams
c = 4.184 J/g°C (specific heat capacity of water)
Initial temperature = 25.0°C
Final temperature = 55.0°C
Step 1: Calculate Δ T , the temperature change: Δ T = Final Temperature − Initial Temperature = 55.0° C − 25.0° C = 30.0° C
Step 2: Substitute the known values into the formula: q = 100.0 g × 4.184 J/g°C × 30.0 ° C
Step 3: Calculate the heat energy: q = 12552 J
Therefore, 12,552 joules of heat are needed to raise the temperature of 100.0 grams of water from 25.0°C to 55.0°C.
