Final answer:
To raise the temperature of a 26-gram sample of water by 35°C, you would need 3800.6 Joules of heat. This calculation is based on the specific heat capacity of water, which is 4.18 J/g°C, and the mass and temperature change of the water sample.
Step-by-step explanation:
To calculate the heat required to raise the temperature of a 26-gram sample of water by 35°C, we need to apply the formula for heat transfer, which is q = mcΔT. Here, q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
For water, the specific heat capacity (c) is commonly given as approximately 4.18 Joules per gram per degree Celsius (J/g°C). The mass of the water (m) is 26 grams and the temperature change (ΔT) is 35°C. The heat absorbed (q) is what we're solving for.
The formula becomes:
q = (26 g)(4.18 J/g°C)(35°C)
→ q = 3800.6 J
This means that 3800.6 Joules of heat are required to raise the temperature of a 26-gram sample of water by 35°C.