400 grams of a metal at 300 C is added to 100 grams of of water at 20 C. Assuming all the heat lost by the metal is gained by the water and that the final temperature or water and metal is 95 C, calculate the specific heat of the metal. Using grams and degrees C, the specific heat of water is 1cal/g C. Just use the calories and don't convert to J.

To solve this problem, we can use the principle of heat transfer, which states that the heat lost by the metal is equal to the heat gained by the water.

The formula for heat transfer is given by:

Q = mcΔT

Where:
- Q is the heat gained or lost (in calories)
- m is the mass (in grams)
- c is the specific heat capacity (in cal/g°C)
- ΔT is the change in temperature (in °C)

Given data:
- Mass of the metal (m1) = 400 grams
- Initial temperature of the metal (T1) = 300°C
- Mass of water (m2) = 100 grams
- Initial temperature of the water (T2) = 20°C
- Final temperature (Tf) = 95°C
- Specific heat capacity of water (c2) = 1 cal/g°C (given)

Now, using the principle of heat transfer, we can set up the equation as follows:

Q lost by metal = Q gained by water

(metal specific heat) * (metal mass) * (metal ΔT) = (water specific heat) * (water mass) * (water ΔT)

(c1) * (m1) * (Tf - T1) = (c2) * (m2) * (Tf - T2)

Since we are looking to find the specific heat of the metal (c1), we can rearrange the equation to solve for it.

c1 = [(c2) * (m2) * (Tf - T2)] / [(m1) * (Tf - T1)]

Now we can substitute the given values into the equation and calculate the specific heat of the metal:

c1 = [(1 cal/g°C) * (100 g) * (95°C - 20°C)] / [(400 g) * (95°C - 300°C)]

Simplifying:

c1 = (75 cal) / (-70,000 cal)

c1 = -0.00107 cal/g°C

Therefore, the specific heat of the metal is approximately -0.00107 cal/g°C.